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Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.2

Question 1.
If 2sin A = 1 = \(\sqrt{2}\) cos B and \(\frac{\pi}{2}\) < A < π, \(\frac{3 \pi}{2}\)
Solution:
Given, 2sin A = 1
∴ sin A = 1/2
we know that,
cos² A = 1 – sin² A = 1 – \(\left(\frac{1}{2}\right)^{2}=1-\frac{1}{4}=\frac{3}{4}\)
∴ cos A = \(\pm \frac{\sqrt{3}}{2}\)
Since \(\frac{\pi}{2}\) < A < π
A lies in the 2nd quadrant.

We know that,
Sin² B = 1 – cos² B = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^{2}\)\(\frac{1}{2}=\frac{1}{2}\)
∴ sin B = \(\pm \frac{1}{\sqrt{2}}\)
Since \(\frac{3 \pi}{2}\) < B < 2π
B lies in the 4th quadrant,

Question 2.
If \(\) and A, B are angles in the second quadran, then prove that 4cosA + 3 cos B = -5
Solution:
Given, \(\frac{\sin \mathrm{A}}{3}=\frac{\sin \mathrm{B}}{4}=\frac{1}{5}\)
∴ sin A = \(\frac{3}{5}\) and sin B = \(\frac{4}{5}\)
We know that,
cos² A = 1 – sin² = 1 – \(\left(\frac{3}{5}\right)^{2}\) = 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ Cos A = ± \([{4}{5}\)
Since A lies in the second quadrant,
cos A < 0
∴ Cos A = –\(\frac{4}{5}\)
Sin B = 4/5
We know that,
cos²B = 1 – sin²B = 1 – \(\left(\frac{4}{5}\right)^{2}=1-\frac{16}{25}=\frac{9}{25}\)
∴ Cos B = ±\(\frac{4}{5}\)
Since B lies in the second quadrant, cos B < 0

Question 3.
If tan θ = \(\frac{1}{2}\), evaluate \(\frac{2 \sin \theta+3 \cos \theta}{4 \cos \theta+3 \sin \theta}\)
Solution:

Question 4.
Eliminate 0 from the following:
i. x = 3sec θ, y = 4tan θ
ii. x = 6cosec θ,y = 8cot θ
iii. x = 4cos θ – 5sin θ, y = 4sin θ + 5cos θ
iv. x = 5 + 6 cosec θ,y = 3 + 8 cot θ
v. x = 3 – 4tan θ,3y = 5 + 3sec θ
Solution:
i. x = 3sec θ, y = 4tan θ
∴ sec θ = \(\frac{x}{3}\) and tan θ= \(\frac{y}{4}\)
We know that,
sec²θ – tan²θ = 1

∴ 16x² – 9y² = 144

ii. x = 6cosec θ and y = 8cot θ
.’. cosec θ = \(\) and cot θ = \(\)
We know that,
cosec² θ – cot² θ =

16x² – 9y² = 576

iii. x = 4cos θ – 5 sin θ … (i)
y = 4sin θ + 5cos θ .. .(ii)
Squaring (i) and (ii) and adding, we get
x² + y² = (4cos θ – 5sin θ)² + (4sin θ + 5cos θ)²
= 16cos²θ – 40 sinθ cosθ + 25 sin²θ + 16 sin² θ + 40sin θ cos θ + 25 cos² θ
= 16(sin² θ + cos² θ) + 25(sin² θ + cos² θ)
= 16(1) + 25(1)
= 41

iv. x = 5 + 6cosec θ andy = 3 + 8cot θ
∴ x – 5 = 6cosec θ and y – 3 = 8cot θ
∴ cosec θ = \(\frac{x-5}{6}\) and cot θ = \(\frac{y-3}{8}\)
We know that,
cosec² θ – cot² θ = 1
∴ \(\left(\frac{x-5}{6}\right)^{2}-\left(\frac{y-3}{8}\right)^{2}\) = 1

v. 2x = 3 – 4tan θ and 3y = 5 + 3sec θ
∴ 2x – 3 = -4tan θ and 3y – 5 = 3sec θ
∴ tan θ = \(\frac{3-2 x}{4}\) and sec θ = \(\frac{3 y-5}{3}\)θ
We know that, sec² θ – tan² θ = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{3-2 x}{4}\right)^{2}\) = 1
∴ \(\left(\frac{3 y-5}{3}\right)^{2}-\left(\frac{2 x-3}{4}\right)^{2}\) = 1

Question 5.
If 2sin² θ + 3sin θ = 0, find the permissible values of cosθ.
Solution:
2sin² θ + 3sin θ = 0
∴ sin θ (2sin θ + 3) = 0
∴ sin θ = 0 or sin θ = \(\frac{-3}{2}\)
Since – 1 ≤ sin θ ≤ 1,
sin θ = 0
\(\sqrt{1-\cos ^{2} \theta}\) = 0 …[ ∵ sin² θ = 1- cos² θ]
∴ 1 – cos² θ = 0
∴ cos² θ = 1
∴ cos θ = ±1 …[∵ – 1 ≤ cos θ ≤ 1]

Question 6.
If 2cos² θ – 11 cos θ + 5 = 0, then find the possible values of cos θ.
Solution:
2cos²θ – 11 cos θ + 5 = 0
∴ 2cos² θ – 10 cos θ – cos θ + 5 = 0
∴ 2cos θ(cos θ – 5) – 1 (cos θ – 5) = 0
∴ (cos θ – 5) (2cos θ – 1) = 0
cos θ – 5 = 0 or 2cos θ – 1 = 0
∴ cos θ = 5 or cos θ = 1/2
Since, -1 ≤ cos θ ≤ 1
∴ cos θ = 1/2

Question 7.
Find the acute angle θ such 2cos² θ = 3sin θ.
Solution:
2cos²0 = 3sin θ
∴ 2(1 – sin² θ) = 3sin θ
∴ 2 – 2sin² θ = 3sin θ
∴ 2sin² θ + 3sin 9-2 = θ
∴ 2sin² θ + 4sin θ – sin θ – 2 = θ
∴ 2sin θ(sin θ + 2) -1 (sin θ + 2) = θ
∴ (sin θ + 2) (2sin θ – 1) = 0
∴ sin θ + 2 = 0 or 2sin θ – 1 = 0
∴ sin θ = -2 or sin θ = 1/2
Since, -1 ≤ sin θ ≤ 1
∴ Sin θ = 1/2
∴ θ = 30° …[ ∵ sin 30 = 1/2]

Question 8.
Find the acute angle 0 such that 5tan2 0 + 3 = 9sec 0.
Solution:
5tan² θ + 3 = 9sec θ
∴ 5(sec² θ – 1) + 3 = 9sec θ
∴ 5sec² θ – 5 + 3 = 9sec θ
∴ 5sec² θ – 9sec θ – 2 = 0
∴ 5sec² θ – 10 sec θ + sec θ – 2 = 0
∴ 5sec θ(sec θ – 2) + 1(sec θ – 2) = 0
∴ (sec θ – 2) (5sec θ + 1) = 0
∴ sec θ – 2 = 0 or 5sec θ + 1 = 0
∴ sec θ = 2 or sec θ = -1/5
Since sec θ ≥ 1 or sec θ ≤ -1,
sec θ = 2
∴ θ = 60° … [ ∵ sec 60° = 2]

Question 9.
Find sin θ such that 3cos θ + 4sin θ = 4.
Solution:
3cos θ + 4sin θ = 4
∴ 3cos θ = 4(1 – sin θ)
Squaring both the sides, we get .
9cos²θ = 16(1 – sin θ)²
∴ 9(1 – sin² θ) = 16(1 + sin² θ – 2sin θ)
∴ 9 – 9sin² θ = 16 + 16sin² θ – 32sin θ
∴ 25sin² θ – 32sin θ + 7 = 0
∴ 25sin² θ – 25sin θ – 7sin θ + 7 = 0
25sin θ (sin θ – 1) – 7 (sin θ – 1) = 0
∴ (sin θ – 1) (25sin θ – 7) = 0
∴ sin θ – 1 = 0 or 25 sin θ – 7 = 0
∴ sin θ = 1 or sin θ = \(\frac{7}{25}\)
Since, -1 ≤ sin θ ≤ 1
∴ sin θ = 1 or \(\frac{7}{25}\)
[Note: Answer given in the textbook is 1. However, as per our calculation it is 1 or \(\frac{7}{25}\).]

Question 10.
If cosec θ + cot θ = 5, then evaluate sec θ.
Solution:
cosec θ + cot θ = 5
∴ \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=5\)
∴ \(\frac{1+\cos \theta}{\sin \theta}=5\)
∴ 1 + cos θ = 5.sin θ
Squaring both the sides, we get
1 + 2 cos θ + cos² θ = 25 sin² θ
∴ cos² θ + 2 cos θ + 1 = 25 (1 – cos² θ)
∴ cos² θ + 2 cos θ + 1 = 25 – 25 cos² θ
∴ 26 cos² θ + 2 cos θ – 24 = 0
∴ 26 cos² θ + 26 cos θ – 24 cos θ – 24 = 0
∴ 26 cos θ (cos θ + 1) – 24 (cos θ + 1) = 0
∴ (cos θ + 1) (26 cos θ – 24) = 0
∴ cos θ + 1 = θ or 26 cos θ – 24 = 0
∴ cos θ = -1 or cos θ = \(\frac{24}{26}=\frac{12}{13}\)
When cos θ = -1, sin θ = 0
∴ cot θ and cosec x are not defined,
∴ cos θ ≠ -1
∴ cos θ = \(\frac{12}{13}\)
∴ sec θ = \(\frac{1}{\cos \theta}=\frac{13}{12}\)
[Note: Answer given in the textbook is -1 or \(\frac{13}{12}\).
However, as per our calculation it is only \(\frac{13}{12}\).]

Question 11.
If cot θ = \(\frac{3}{4}\) and π < θ < \(\frac{3 \pi}{2}\), then find the value of 4 cosec θ + 5 cos θ.
Solution:
We know that,
cosec²θ = 1 + cot² θ = \(\left(\frac{3}{4}\right)^{2}\) = 1 + \(\frac{9}{16}\)
∴ cosec² θ = \(\frac{25}{16}\)
∴ cosec θ = \(\pm \frac{5}{4}\)
Since π < θ < \(\frac{3 \pi}{2}\)
θ lies in the third quadrant.
∴ cosec θ < 0
∴ cosec θ = –\(\frac{5}{4}\)
cot θ = \(\frac{3}{4}\)
tan θ = \(\frac{1}{\cot \theta}=\frac{4}{3}\)
We know that,
sec² θ = 1 + tan² θ = 1 + \(\left(\frac{4}{3}\right)^{2}\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ sec θ = ±\(\frac{5}{3}\)
Since θ lies in the third quadrant,
sec θ < 0
∴ sec θ = –\(\frac{5}{3}\)
cos θ = \(\frac{1}{\sec \theta}=\frac{-3}{5}\)
∴ 4cosec θ + 5cos θ
= \(4\left(-\frac{5}{4}\right)+5\left(-\frac{3}{5}\right)\)
= -5 – 3 = -8
[Note: The question has been modified.]

Question 12.
Find the Cartesian co-ordinates of points whose polar co-ordinates are:
i. (3, 90°) ii. (1, 180°)
Solution:
i. (r, θ) = (3, 90°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 3cos 90° and y = 3sin 90°
∴ x = 3(0) = 0 and y = 3(1) = 3
∴ the required cartesian co-ordinates are (0, 3).

ii. (r, θ) = (1, 180°)
Using x = r cos θ and y = r sin θ, where (x, y) are the required cartesian co-ordinates, we get
x = 1(cos 180°) and y = 1(sin 180°)
∴ x = -1 and y = 0
∴ the required cartesian co-ordinates are (-1, 0).

Question 13.
Find the polar co-ordinates of points whose Cartesian co-ordinates are:
1. (5, 5) ii. (1, \(\sqrt{3}\))
ii. (-1, -1) iv. (-\(\sqrt{3}\), 1)
Solution:
i. (x, y) = (5, 5)
∴ r = \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{25+25}\)
\(=\sqrt{50}=5 \sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{5}{5}\) = 1
Since the given point lies in the 1st quadrant,
θ = 45° …[∵ tan 45° = 1]
∴ the required polar co-ordinates are (\(5 \sqrt{2}\), 45°).

ii. (x, y) = ( 1, \(\sqrt{3}\))
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}\)
Since the given point lies in the 1st quadrant,
θ = 60° …[∵ tan 60° = \(\sqrt{3}\)]
∴ the required polar co-ordinates are (2, 60°).

iii. (x, y) = (-1, -1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{1+1}=\sqrt{2}\)
tan θ = \(\frac{y}{x}=\frac{-1}{-1}=1\)
∴ tan θ = tan \(\frac{\pi}{4}\)
Since the given point lies in the 3rd quadrant,
tan θ = tan \(\left(\pi+\frac{\pi}{4}\right)\) …[∵ tan (n + x) = tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{4}\right)\)
∴ θ = \(\frac{5 \pi}{4}\) = 225°
∴ the required polar co-ordinates are (\(\sqrt{2}\), 225°).

iv. (x, y) = (-\(\sqrt{3}\) , 1)
∴ r = \(\sqrt{x^{2}+y^{2}}=\sqrt{3+1}=\sqrt{4}=2\)
tan θ = \(\frac{y}{x}=\frac{1}{-\sqrt{3}}\) = -tan \(\frac{\pi}{6}\)
Since the given point lies in the 2nd quadrant,
tan θ = tan \(\left(\pi-\frac{\pi}{6}\right)\) …[∵ tan (π – x) = – tanx]
∴ tan θ = tan \(\left(\frac{5 \pi}{6}\right)\)
∴ θ = \(\frac{5 \pi}{6}\) = 150°
∴ the required polar co-ordinates are (2, 150°)

Question 14.
Find the values of:
i. sin\(\frac{19 \pi^{e}}{3}\)
ii. cos 1140°
iii. cot \(\frac{25 \pi^{e}}{3}\)
Solution:
i. We know that sine function is periodic with period 2π.
sin \(\frac{19 \pi}{3}\) = sin \(\left(6 \pi+\frac{\pi}{3}\right)\) = sin \(\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

ii. We know that cosine function is periodic with period 2π.
cos 1140° = cos (3 × 360° + 60°)
= cos 60° = \(\frac {1}{2}\)

iii. We know that cotangent function is periodic with period π.
cot \(\frac{25 \pi}{3}\) = cot \(\left(8 \pi+\frac{\pi}{3}\right)\) = cot \(\frac{\pi}{3}\) = \(\frac{1}{\sqrt{3}}\)
dhana work.txt
Displaying dhana work.txt.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.1

Question 1.
The vector \(\bar{a}\) is directed due north and \(|\bar{a}|\) = 24. The vector \(\bar{b}\) is directed due west and \(|\bar{b}|\) = 7. Find \(|\bar{a}+\bar{b}|\).
Solution:

Let \(\overline{\mathrm{AB}}\) = \(\bar{a}\), \(\overline{\mathrm{BC}}\) = \(\bar{b}\)
Then \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\) = a + b
Given : \(|\bar{a}|\) = \(|\overline{\mathrm{AB}}|\) = l(AB) = 24 and
\(|\bar{b}|\) = \(|\overline{\mathrm{BC}}|\) = l(BC) = 7
∴ ∠ABC = 90°
∴ [l(AC)]² = [l(AB)]² + [l(BC)]²
= (24)² + (7)² = 625
∴ l(AC) = 25 ∴ \(|\overline{\mathrm{AC}}|\) = 25
∴ \(|\bar{a}+\bar{b}|\) = \(|\overline{\mathrm{AC}}|\) = 25.

Question 2.
In the triangle PQR, \(\overline{\mathrm{PQ}}\) = 2\(\bar{a}\) and \(\overline{\mathrm{QR}}\) = 2\(\bar{b}\). The mid-point of PR is M. Find following vectors in terms of \(\bar{a}\) and \(\bar{b}\).
(i) \(\overline{\mathrm{PR}}\)
Solution:

Given : \(\overline{\mathrm{PQ}}\) = 2\(\bar{a}\), \(\overline{\mathrm{QR}}\) = 2\(\bar{b}\)
(i) \(\overline{\mathrm{PR}}\) = \(\overline{\mathrm{PQ}}\) + \(\overline{\mathrm{QR}}\)
= 2\(\bar{a}\) + 2\(\bar{a}\).

(ii) \(\overline{\mathrm{PM}}\)
Solution:
∵ M is the midpoint of PR
∴ \(\overline{\mathrm{PM}}\) = \(\frac{1}{2} \overline{\mathrm{PR}}\) = \(\frac{1}{2}\)[2\(\bar{a}\) + 2\(\bar{b}\)]
= \(\bar{a}\) + \(\bar{b}\).

(iii) \(\overline{\mathrm{QM}}\)
Solution:
\(\overline{\mathrm{RM}}\) = \(\frac{1}{2}(\overline{\mathrm{RP}})\) = \(-\frac{1}{2} \overline{\mathrm{PR}}\) = \(-\frac{1}{2}\)(2\(\bar{a}\) + 2\(\bar{b}\))
= –\(\bar{a}\) – \(\bar{b}\)
∴ \(\overline{\mathrm{QM}}\) = \(\overline{\mathrm{QR}}\) + \(\overline{\mathrm{RM}}\)
= 2\(\bar{b}\) – \(\bar{a}\) – \(\bar{b}\)
= \(\bar{b}\) – \(\bar{a}\).

Question 3.
OABCDE is a regular hexagon. The points A and B have position vectors \(\bar{a}\) and \(\bar{b}\) respectively, referred to the origin O. Find, in terms of \(\bar{a}\) and \(\bar{b}\) the position vectors of C, D and E.
Solution:

Given : \(\overline{\mathrm{OA}}\) = \(\bar{a}\), \(\overline{\mathrm{OB}}\) = \(\bar{a}\) Let AD, BE, OC meet at M.
Then M bisects AD, BE, OC.
\(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{AO}}\) + \(\overline{\mathrm{OB}}\) = –\(\overline{\mathrm{OA}}\) + \(\overline{\mathrm{OB}}\) = –\(\bar{a}\) + \(\bar{b}\) = \(\bar{b}\) – \(\bar{a}\)
∵ OABM is a parallelogram

Hence, the position vectors of C, D and E are 2\(\bar{b}\) – 2\(\bar{a}\), 2\(\bar{b}\) – 3\(\bar{a}\) and \(\bar{b}\) – 2\(\bar{a}\) respectively.

Question 4.
If ABCDEF is a regular hexagon, show that \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{AC}}\) + \(\overline{\mathrm{AD}}\) + \(\overline{\mathrm{AE}}\) + \(\overline{\mathrm{AF}}\) = 6\(\overline{\mathrm{AO}}\), where O is the center of the hexagon.
Solution:

ABCDEF is a regular hexagon.
∴ \(\overline{\mathrm{AB}}\) = \(\overline{\mathrm{ED}}\) and \(\overline{\mathrm{AF}}\) = \(\overline{\mathrm{CD}}\)
∴ by the triangle law of addition of vectors,

Question 5.
Check whether the vectors \(2 \hat{i}+2 \hat{j}+3 \hat{k}\), + \(-3 \hat{i}+3 \hat{j}+2 \hat{k}\), + \(3 \hat{i}+4 \hat{k}\) form a triangle or not.
Solution:
Let, if possible, the three vectors form a triangle ABC
with \(\overline{A B}\) = \(2 \hat{i}+2 \hat{j}+3 \hat{k}\), \(\overline{B C}\) = \(3 \hat{i}+3 \hat{j}+2 \hat{k}\), \(\overline{A C}\) = \(3 \hat{i}+4 \hat{k}\)
Now, \(\overline{A B}\) + \(\overline{B C}\)
= \((2 \hat{i}+2 \hat{j}+3 \hat{k})\) + \((-3 \hat{i}+3 \hat{j}+2 \hat{k})\)
= \(-\hat{i}+5 \hat{j}+5 \hat{k} \neq 3 \hat{i}+4 \hat{k}\) = \(\overline{\mathrm{AC}}\)
Hence, the three vectors do not form a triangle.

Question 6.
In the figure 5.34 express \(\bar{c}\) and \(\bar{d}\) in terms of \(\bar{a}\) and \(\bar{b}\). Find a vector in the direction of \(\bar{a}\) = \(\hat{i}-2 \hat{j}\) that has magnitude 7 units.

Solution:
\(\overline{\mathrm{PQ}}\) = \(\overline{\mathrm{PS}}\) + \(\overline{\mathrm{SQ}}\)
∴ \(\bar{a}\) = \(\bar{c}\) – \(\bar{d}\) … (1)
\(\overline{\mathrm{PR}}\) = \(\overline{\mathrm{PS}}\) + \(\overline{\mathrm{SR}}\)
∴ \(\bar{b}\) = \(\bar{c}\) + \(\bar{d}\) … (2)
Adding equations (1) and (2), we get
\(\bar{a}\) + \(\bar{b}\) = (\(\bar{c}\) – \(\bar{d}\)) + (\(\bar{c}\) + \(\bar{d}\)) = 2\(\bar{c}\)

Question 7.
Find the distance from (4, -2, 6) to each of the following :
(a) The XY-plane
Solution:
Let the point A be (4, -2, 6).
Then,
The distance of A from XY-plane = |z| = 6

(b) The YZ-plane
Solution:
The distance of A from YZ-plane = |x| = 4

(c) The XZ-plane
Solution:
The distance of A from ZX-plane = |y| = 2

(d) The X-axis
Solution:
The distance of A from X-axis
= \(\sqrt{y^{2}+z^{2}}\) = \(\sqrt{(-2)^{2}+6^{2}}\) = \(\sqrt{40}\) = \(2 \sqrt{10}\)

(e) The Y-axis
Solution:
The distance of A from Y-axis
= \(\sqrt{z^{2}+x^{2}}\) = \(\sqrt{6^{2}+4^{2}}\) = \(\sqrt{52}\) = \(2 \sqrt{13}\)

(f) The Z-axis
Solution:
The distance of A from Z-axis
= \(\sqrt{x^{2}+y^{2}}\) = \(\sqrt{4^{2}+(-2)^{2}}\) = \(\sqrt{20}\) = \(2 \sqrt{5}\)

Question 8.
Find the coordinates of the point which is located :
(a) Three units behind the YZ-plane, four units to the right of the XZ-plane and five units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located 3 units behind the YZ- j plane, 4 units to the right of XZ-plane and 5 units , above the XY-plane,
x = -3, y = 4 and z = 5
Hence, coordinates of the required point are (-3, 4, 5)

(b) In the YZ-plane, one unit to the right of the XZ-plane and six units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located in the YZ plane, x = 0. Also, the point is one unit to the right of XZ-plane and six units above the XY-plane.
∴ y = 1, z = 6.
Hence, coordinates of the required point are (0, 1, 6).

Question 9.
Find the area of the triangle with vertices (1, 1, 0), (1, 0, 1) and (0, 1, 1).
Solution:
Let A = (1, 1, 0), B = (1, 0, 1), C = (0, 1, 1)

Question 10.
If \(\overline{\mathrm{AB}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\) and initial point A ≡ (1, 5, ,0). Find the terminal point B.
Solution:
Let \(\bar{a}\) and \(\bar{b}\) be the position vectors of A and B.
Given : A = (1, 5, 0) .’. \(\bar{a}\) = \(\hat{i}+5 \hat{j}\)
Now, \(\overline{\mathrm{AB}}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
∴ \(\bar{b}\) – \(\bar{a}\) = \(2 \hat{i}-4 \hat{j}+7 \hat{k}\)
∴ \(\bar{b}\) = \((2 \hat{i}-4 \hat{j}+7 \hat{k})\) + \(\bar{a}\)
= \((2 \hat{i}-4 \hat{j}+7 \hat{k})\) + \((\hat{i}+5 \hat{j})\)
= \(3 \hat{i}+\hat{j}+7 \hat{k}\)
Hence, the terminal point B = (3, 1, 7).

Question 11.
Show that the following points are collinear :
(i) A (3, 2, -4), B (9, 8, -10), C (-2, -3, 1).
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) be the position vectors of the points.
A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.

∴ \(\overline{\mathrm{BC}}\) is a non-zero scalar multiple of \(\overline{\mathrm{AB}}\)
∴ they are parallel to each other.
But they have the point B in common.
∴ \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AB}}\) are collinear vectors.
Hence, the points A, B and C are collinear.

(ii) P (4, 5, 2), Q (3, 2, 4), R (5, 8, 0).
Solution:
Let \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) be the position vectors of the points.
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.

= 2.\(\overline{\mathrm{AB}}\) …[By (1)]
∴ \(\overline{\mathrm{BC}}\) is a non-zero scalar multiple of \(\overline{\mathrm{AB}}\)
∴ they are parallel to each other.
But they have the point B in common.
∴ \(\overline{\mathrm{BC}}\) and \(\overline{\mathrm{AB}}\) are collinear vectors.
Hence, the points A, B and C are collinear.

Question 12.
If the vectors \(2 \hat{i}-q \hat{j}+3 \hat{k}\) and \(4 \hat{i}-5 \hat{j}+6 \hat{k}\) are collinear, then find the value of q.
Solution:
The vectors \(2 \hat{i}-q \hat{j}+3 \hat{k}\) and \(4 \hat{i}-5 \hat{j}+6 \hat{k}\) are collinear
∴ the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) are proportional

Question 13.
Are the four points A(1, -1, 1), B(-1, 1, 1), C(1, 1, 1) and D(2, -3, 4) coplanar? Justify your answer.
Solution:
The position vectors \(\bar{a}\), \(\bar{b}\), \(\bar{c}\), \(\bar{d}\) of the points A, B, C, D are

By equality of vectors,
y = -2 ….(1)
2x – 2y = 2 … (2)
3y = 0 … (3)
From (1), y = -2
From (3), y = 0 This is not possible.
Hence, the points A, B, C, D are not coplanar.

Question 14.
Express \(-\hat{i}-3 \hat{j}+4 \hat{k}\) as linear combination of the vectors \(2 \hat{i}+\hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+3 \hat{k}\) and \(3 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:

By equality of vectors,
2x + 2y + 3 = -1
x – y + z = -3
-4x + 3y – 2z = 4
We have to solve these equations by using Cramer’s Rule
D = \(\left|\begin{array}{rrr}
2 & 2 & 3 \\
1 & -1 & 1 \\
-4 & 3 & -2
\end{array}\right|\)
= 2(2 – 3) – 2(-2 + 4) + 3(3 – 4)
= -2 – 4 – 3 = -9 ≠ 0

= 2(-4 + 9) – 2(4 – 12) – 1(3 – 4)
= 10 + 16 + 1 = 27

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 5 Vectors Ex 5.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.4

Question 1.
If \(\bar{a}\) = \(2 \hat{i}+3 \hat{j}-\hat{k}\), \(\bar{b}\) = \(\hat{i}-4 \hat{j}+2 \hat{k}\) find (\(\bar{a}\) + \(\bar{b}\)) × (\(\bar{a}\) – \(\bar{b}\))
Solution:

Question 2.
Find a unit vector perpendicular to the vectors \(\hat{j}+2 \hat{k}\) and \(\hat{i}+\hat{j}\).
Solution:
Let \(\bar{a}\) = \(\hat{j}+2 \hat{k}\), \(\bar{b}\) = \(\hat{i}+\hat{j}\)

Question 3.
If \(\bar{a} \cdot \bar{b}\) = \(\sqrt {3}\) and \(\bar{a} \times \bar{b}\) = \(2 \hat{i}+\hat{j}+2 \hat{k}\), find the angle between \(\bar{a}\) and \(\bar{b}\).
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\)

∴ θ = 60°.

Question 4.
If \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\), find a vector of magnitude 5 perpendicular to both \(\bar{a}\) and \(\bar{b}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}\) = \(\hat{i}-2 \hat{j}+\hat{k}\)

∴ unit vectors perpendicular to both the vectors \(\bar{a}\) and \(\bar{b}\).

∴ required vectors of magnitude 5 units
= ±\(\frac{5}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)

Question 5.
Find
(i) \(\bar{u}\)∙\(\bar{v}\) if \(|\bar{u}|\) = 2, \(|\vec{v}|\) = 5, \(|\bar{u} \times \bar{v}|\) = 8
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(|\bar{u} \times \bar{v}|\) = 8 gives
\(|\bar{u}||\bar{v}|\) sin θ = 8
∴ 2 × 5 × sin θ = 8

(ii) \(|\bar{u} \times \bar{v}|\) if \(|\bar{u}|\) = 10, \(|\vec{v}|\) = 2, \(\bar{u} \cdot \bar{v}\) = 12
Solution:
Let θ be the angle between \(\bar{u}\) and \(\bar{v}\).
Then \(\bar{u} \cdot \bar{v}\) = 12 gives
\(|\bar{u} \| \bar{v}|\)cos θ = 12
∴ 10 × 2 × cos θ = 12

Question 6.
Prove that 2(\(\bar{a}\) – \(\bar{b}\)) × 2(\(\bar{a}\) + \(\bar{b}\)) = 8(\(\bar{a}\) × \(\bar{b}\))
Solution:

Question 7.
If \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), and \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\), verify that \(\bar{a}\) × (\(\bar{b}\) + \(\bar{c}\)) = \(\bar{a}\) × \(\bar{b}\) + \(\bar{a}\) × \(\bar{c}\)
Solution:
Given : \(\bar{a}\) = \(\hat{i}-2 \hat{j}+3 \hat{k}\), \(\bar{b}\) = \(4 \hat{i}-3 \hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{i}-\hat{j}+2 \hat{k}\)

Question 8.
Find the area of the parallelogram whose adjacent sides are the vectors \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\) and \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\).
Solution:
Given : \(\bar{a}\) = \(2 \hat{i}-2 \hat{j}+\hat{k}\), \(\bar{b}\) = \(\hat{i}-3 \hat{j}-3 \hat{k}\)

Area of the parallelogram whose adjacent sides are \(\bar{a}\) and \(\bar{b}\) is \(|\bar{a} \times \bar{b}|\) =\(\sqrt {146}\) sq units.

Question 9.
Show that vector area of a quadrilateral ABCD is \(\frac{1}{2}\) (\(\overline{A C}\) × \(\overline{B D}\)), where AC and BD are its diagonals.
Solution:
Let ABCD be a parallelogram.
Then \(\overline{\mathrm{AC}}\) = \(\overline{\mathrm{AB}}\) + \(\overline{\mathrm{BC}}\)

Question 10.
Find the area of parallelogram whose diagonals are determined by the vectors \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), and \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)
Solution:
Given: \(\bar{a}\) = \(3 i-\hat{j}-2 \hat{k}\), \(\bar{b}\) = \(-\hat{i}+3 \hat{j}-3 \hat{k}\)

Question 11.
If \(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors such that \(\bar{a} \times \bar{b}=\bar{c} \times \bar{d}\) and \(\bar{a} \times \bar{c}=\bar{b} \times \bar{d}\), prove that \(\bar{a}\) – \(\bar{d}\) is parallel to \(\bar{b}\) – \(\bar{c}\).
Solution:
\(\bar{a}\), \(\bar{b}\), \(\bar{c}\) and \(\bar{d}\) are four distinct vectors

Question 12.
If \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\) and, \(\bar{c}\) = \(\hat{j}-\hat{k}\), find a vector \(\bar{b}\) satisfying \(\bar{a}\) × \(\bar{b}\) = \(\bar{c}\) and \(\bar{a} \cdot \bar{b}\) = 3
Solution:
Given \(\bar{a}\) = \(\hat{i}+\hat{j}+\hat{k}\), \(\bar{c}\) = \(\hat{j}-\hat{k}\)

By equality of vectors,
z – y = 0 ….(2)
x – z = 1 ……(3)
y – x = -1 ……(4)
From (2), y = z.
From (3), x = 1 + z
Substituting these values of x and y in (1), we get
1 + z + z + z = 3 ∴ z = \(\frac{2}{3}\)
∴ y = z = \(\frac{2}{3}\)
∴ x = 1 + z =1 + \(\frac{2}{3}=\frac{5}{3}\)

Question 13.
Find \(\bar{a}\), if \(\bar{a} \times \hat{i}+2 \bar{a}-5 \hat{j}=\overline{0}\).
Solution:

By equality of vectors
2x = 0 i.e. x = 0
2y + z – 5 = 0 … (1)
2z – y = 0 … (2)
From (2), y = 2z
Substituting y = 2z in (1), we get
4z + z = 5 ∴ z = 1
∴ y = 2z = 2(1) = 2
∴ x = 0, y = 2, z = 1
∴ \(\bar{a}=2 \hat{j}+\hat{k}\)

Question 14.
If \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) and \(\bar{a} \cdot \bar{b}\) < 0, then find the angle between \(\bar{a}\) and \(\bar{b}\)
Solution:
Let θ be the angle between \(\bar{a}\) and \(\bar{b}\).
Then \(|\bar{a} \cdot \bar{b}|\) = \(|\bar{a} \times \bar{b}|\) gives

Hence, the angle between \(\bar{a}\) and \(\bar{b}\) is \(\frac{3 \pi}{4}\).

Question 15.
Prove by vector method that sin (α + β) = sinα∙cosβ+cosα∙sinβ.
Solution:

Let ∠XOP and ∠XOQ be in standard position and m∠XOP = -α, m∠XOQ = β.
Take a point A on ray OP and a point B on ray OQ such that
OA = OB = 1.
Since cos (-α) = cos α
and sin (-α) = -sin α,
A is (cos (-α), sin (-α)),
i.e. (cos α, – sin α)
B is (cos β, sin β)

The angle between \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is α + β.
Also \(\overline{\mathrm{OA}}\), \(\overline{\mathrm{OB}}\) lie in the XY-plane.
∴ the unit vector perpendicular to \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) is \(\bar{k}\).
∴ \(\overline{\mathrm{OA}}\) × \(\overline{\mathrm{OB}}\) = [OA∙OB sin (α + β)]\(\bar{k}\)
= sin(α + β)∙\(\bar{k}\) …(2)
∴ from (1) and (2),
sin (α + β) = sin α cos β + cos α sin β.

Question 16.
Find the direction ratios of a vector perpendicular to the two lines whose direction ratios are
(i) -2, 1, -1 and -3, -4, 1
Solution:
Let a, b, c be the direction ratios of the vector which is perpendicular to the two lines whose direction ratios are -2, 1, -1 and -3, -4, 1
∴ -2a + b – c = 0 and -3a – 4b + c = 0

∴ the required direction ratios are -3, 5, 11
Alternative Method:
Let \(\bar{a}\) and \(\bar{b}\) be the vectors along the lines whose direction ratios are -2, 1, -1 and -3, -4, 1 respectively.
Then \(\bar{a}\) = \(-2 \hat{i}+\hat{j}-\hat{k}\) and \(\bar{b}\) = \(-3 \hat{i}-4 \hat{j}+\hat{k}\)
The vector perpendicular to both \(\bar{a}\) and \(\bar{b}\) is given by

Hence, the required direction ratios are -3, 5, 11.

(ii) 1, 3, 2 and -1, 1, 2
Solution:

Question 17.
Prove that two vectors whose direction cosines are given by relations al + bm + cn = 0 and fmn + gnl + hlm = 0 are perpendicular if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0
Solution:
Given, al + bm + cn = 0 …(1)
and fmn + gnl + hlm = 0 …..(2)
From (1), n = \(-\left(\frac{a l+b m}{c}\right)\) …..(3)
Substituting this value of n in equation (2), we get
(fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0
∴ -(aflm + bfm² + agl² + bglm) + chlm = 0
∴ agl² + (af + bg – ch)lm + bfm² = 0 … (4)
Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get
n = 0, which is not possible as l² + m² + n² = 1.
Let us take m # 0.
Dividing equation (4) by m², we get
ag\(\left(\frac{l}{m}\right)^{2}\) + (af + bg – ch)\(\left(\frac{l}{m}\right)\) + bf = 0 … (5)
This is quadratic equation in \(\left(\frac{l}{m}\right)\).
If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of the two lines given by the equation (1) and (2), then \(\frac{l_{1}}{m_{1}}\) and \(\frac{l_{2}}{m_{2}}\) are the roots of the equation (5).
From the quadratic equation (5), we get

i.e. if \(\frac{f}{a}+\frac{g}{b}+\frac{h}{c}\) = 0.

Question 18.
If A(1, 2, 3) and B(4, 5, 6) are two points, then find the foot of the perpendicular from the point B to the line joining the origin and point A.
Solution:

Let M be the foot of the perpendicular drawn from B to the line joining O and A.
Let M = (x, y, z)
OM has direction ratios x – 0, y – 0, z – 0 = x, y, z
OA has direction ratios 1 – 0, 2 – 0, 3 – 0 = 1, 2, 3
But O, M, A are collinear.
∴ \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}\) = k …(Let)
∴ x = k, y = 2k, z = 3k
∴ M = (k, 2k, 3k)
∵ BM has direction ratios
k – 4, 2k – 5, 3k – 6
BM is perpendicular to OA
∴ (l)(k – 4) + 2(2k – 5) + 3(3k – 6)
∴ = k – 4 + 4k – 10 + 9k – 18 = 0
∴ 14k = 32
∴ k = \(\frac{16}{7}\)
∴ M = (k, 2k, 3k) = (\(\frac{16}{7}, \frac{32}{7}, \frac{48}{7}\))

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Linear Programming Ex 7.4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Linear Programming Ex 7.4

Question 1.
Maximize : z = 11x + 8y subject to x ≤ 4, y ≤ 6,
x + y ≤ 6, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and ED whose equations are x = 4, y = 6 and x + y = 6 respectively.


The feasible region is shaded portion OAPDO in the graph.
The vertices of the feasible region are O (0, 0), A (4, 0), P and D (0, 6)
P is point of intersection of lines x + y = 6 and x = 4.
Substituting x = 4 in x + y = 6, we get
4 + y = 6 ∴ y = 2 ∴ P is (4, 2).
∴ the corner points of feasible region are O (0, 0), A (4, 0), P(4, 2) and D(0 ,6).
The values of the objective function z = 11x + 8y at these vertices are
z (O) = 11(0) + 8(0) = 0 + 0 = 0
z(a) = 11(4) + 8(0) = 44 + 0 = 44
z (P) = 11(4) + 8(2) = 44 + 16 = 60
z (D) = 11(0) + 8(2) = 0 + 16 = 16
∴ z has maximum value 60, when x = 4 and y = 2.

Question 2.
Maximize : z = 4x + 6y subject to 3x + 2y ≤ 12,
x + y ≥ 4, x, y ≥ 0.
Solution:
First we draw the lines AB and AC whose equations are 3x + 2y = 12 and x + y = 4 respectively.


The feasible region is the ∆ABC which is shaded in the graph.
The vertices of the feasible region (i.e. corner points) are A (4, 0), B (0, 6) and C (0, 4).
The values of the objective function z = 4x + 6y at these vertices are
z(a) = 4(4) + 6(0) = 16 + 0 = 16
z(B) = 4(0)+ 6(6) = 0 + 36 = 36
z(C) = 4(0) + 6(4) = 0 + 24 = 24
∴ has maximum value 36, when x = 0, y = 6.

Question 3.
Maximize : z = 7x + 11y subject to 3x + 5y ≤ 26
5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 3x + 5y = 26 and 5x + 3y = 30 respectively.


The feasible region is OCPBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (6, 0), p and B(0, \(\frac{26}{5}\))
The vertex P is the point of intersection of the lines
3x + 5y = 26 … (1)
and 5x + 3y = 30 … (2)
Multiplying equation (1) by 3 and equation (2) by 5, we get
9x + 15y = 78
and 25x + 15y = 150
On subtracting, we get
16x = 72 ∴ x = \(\frac{72}{16}=\frac{9}{2}\) = 4.5
Substituting x = 4.5 in equation (2), we get
5(4.5) + 3y = 30
22.5 + 3y = 30
∴ 3y = 7.5 ∴ y = 2.5
∴ P is (4.5, 2.5)
The values of the objective function z = 7x + 11y at these corner points are
z (O) = 7(0) + 11(0) = 0 + 0 = 0
z (C) = 7(6) + 11(0) = 42 + 0 = 42
z (P) = 7(4.5) + 11 (2.5) = 31.5 + 27.5 = 59.0 = 59
z(B) = 7(0) + 11\(\left(\frac{26}{5}\right)=\frac{286}{5}\) = 57.2
∴ z has maximum value 59, when x = 4.5 and y = 2.5.

Question 4.
Maximize : z = 10x + 25y subject to 0 ≤ x ≤ 3,
0 ≤ y ≤ 3, x + y ≤ 5 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x = 3, y = 3 and x + y = 5 respectively.


The feasible region is OAPQDO which is shaded in the i graph.
The vertices of the feasible region are O (0, 0), A (3, 0), P, Q and D(0, 3).
t P is the point of intersection of the lines x + y = 5 and x = 3.
Substituting x = 3 in x + y = 5, we get
3 + y = 5 ∴ y = 2
∴ P is (3, 2)
Q is the point of intersection of the lines x + y = 5 and y = 3
Substituting y = 3 in x + y = 5, we get
x + 3 = 5 ∴ x = 2
∴ Q is (2, 3)
The values of the objective function z = 10x + 25y at these vertices are
z(O) = 10(0) + 25(0) = 0 + 0 = 0
z(a) = 10(3) + 25(0) = 30 + 0 = 30
z(P) = 10(3) + 25(2) = 30 + 50 = 80
z(Q) = 10(2) + 25(3) = 20 + 75 = 95
z(D) = 10(0)+ 25(3) = 0 + 75 = 75
∴ z has maximum value 95, when x = 2 and y = 3.

Question 5.
Maximize : z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21,
x + y ≤ 9, x ≥ 0, y ≥ 0 also find maximum value of z.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 4y = 24, 3x + y = 21 and x + y = 9 respectively.


The feasible region is OCPQBO which is shaded in the graph.
The vertices of the feasible region are O (0, 0), C (7, 0), P, Q and B (0, 6).
P is the point of intersection of the lines
3x + y = 21 … (1)
and x + y = 9 … (2)
On subtracting, we get 2x = 12 ∴ x = 6
Substituting x = 6 in equation (2), we get
6 + y = 9 ∴ y = 3
∴ P = (6, 3)
Q is the point of intersection of the lines
x + 4y = 24 … (3)
and x + y = 9 … (2)
On subtracting, we get
3y = 15 ∴ y = 5
Substituting y = 5 in equation (2), we get
x + 5= 9 ∴ x = 4
∴ Q = (4, 5)
∴ the corner points of the feasible region are 0(0,0), C(7, 0), P (6, 3), Q (4, 5) and B (0, 6).
The values of the objective function 2 = 3x + 5y at these corner points are
z(O) = 3(0)+ 5(0) = 0 + 0 = 0
z(C) = 3(7) + 5(0) = 21 + 0 = 21
z(P) = 3(6) + 5(3) = 18 + 15 = 33
z(Q) = 3(4) + 5(5) = 12 + 25 = 37
z(B) = 3(0)+ 5(6) = 0 + 30 = 30
∴ z has maximum value 37, when x = 4 and y = 5.

Question 6.
Minimize : z = 7x + y subject to 5x + y ≥ 5, x + y ≥ 3,
x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB and CD whose equations are 5x + y = 5 and x + y = 3 respectively.


The feasible region is XCPBY which is shaded in the graph.
The vertices of the feasible region are C (3, 0), P and B (0, 5).
P is the point of the intersection of the lines
5x + y = 5
and x + y = 3
On subtracting, we get
4x = 2 ∴ x = \(\frac{1}{2}\)
Substituting x = \(\frac{1}{2}\) in x + y = 3, we get
\(\frac{1}{2}\) + y = 3
∴ y = \(\frac{5}{2}\) ∴ P = \(\left(\frac{1}{2}, \frac{5}{2}\right)\)
The values of the objective function z = 7x + y at these vertices are
z(C) = 7(3) + 0 = 21
z(B) = 7(0) + 5 = 5
∴ z has minimum value 5, when x = 0 and y = 5.

Question 7.
Minimize : z = 8x + 10y subject to 2x + y ≥ 7, 2x + 3y ≥ 15,
y ≥ 2, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are 2x + y = 7, 2x + 3y = 15 and y = 2 respectively.


The feasible region is EPQBY which is shaded in the graph. The vertices of the feasible region are P, Q and B(0,7). P is the point of intersection of the lines 2x + 3y = 15 and y = 2.
Substituting y – 2 in 2x + 3y = 15, we get 2x + 3(2) = 15
∴ 2x = 9 ∴ x = 4.5 ∴ P = (4.5, 2)
Q is the point of intersection of the lines
2x + 3y = 15 … (1)
and 2x + y = 7 … (2)
On subtracting, we get
2y = 8 ∴ y = 4
∴ from (2), 2x + 4 = 7
∴ 2x = 3 ∴ x = 1.5
∴ Q = (1.5, 4)
The values of the objective function z = 8x + 10y at these vertices are
z(P) = 8(4.5) + 10(2) = 36 + 20 = 56
z(Q) = 8(1.5) + 10(4) = 12 + 40 = 52
z(B) = 8(0) +10(7) = 70
∴ z has minimum value 52, when x = 1.5 and y = 4

Question 8.
Minimize : z = 6x + 21y subject to x + 2y ≥ 3, x + 4y ≥ 4,
3x + y ≥ 3, x ≥ 0, y ≥ 0.
Solution:
First we draw the lines AB, CD and EF whose equations are x + 2y = 3, x + 4y = 4 and 3x + y = 3 respectively.


The feasible region is XCPQFY which is shaded in the graph.
The vertices of the feasible region are C (4, 0), P, Q and F(0, 3).
P is the point of intersection of the lines x + 4y = 4 and x + 2y = 3
On subtracting, we get
2y = 1 ∴ y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in x + 2y = 3, we get
x + 2\(\left(\frac{1}{2}\right)\) = 3
∴ x = 2
∴ P = (2, \(\frac{1}{2}\))
Q is the point of intersection of the lines
x + 2y = 3 … (1)
and 3x + y = 3 ….(2)
Multiplying equation (1) by 3, we get 3x + 6y = 9
Subtracting equation (2) from this equation, we get
5y = 6
∴ y = \(\frac{6}{5}\)
∴ from (1), x + 2\(\left(\frac{6}{5}\right)\) = 3
∴ x = 3 – \(\frac{12}{5}=\frac{3}{5}\)
Q ≡ \(\left(\frac{3}{5}, \frac{6}{5}\right)\)
The values of the objective function z = 6x + 21y at these vertices are
z(C) = 6(4) + 21(0) = 24
z(P) = 6(2) + 21\(\left(\frac{1}{2}\right)\)
= 12 + 10.5 = 22.5
z(Q)= 6\(\left(\frac{3}{5}\right)\) + 21\(\left(\frac{6}{5}\right)\)
= \(\frac{18}{5}+\frac{126}{5}=\frac{144}{5}\) = 28.8
2 (F) = 6(0) + 21(3) = 63
∴ z has minimum value 22.5, when x = 2 and y = \(\frac{1}{2}\).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Miscellaneous Exercise 4

I : Choose correct alternatives.
Question 1.
If the equation 4x² + hxy + y² = 0 represents two coincident lines, then h = _________.
(A) ± 2
(B) ± 3
(C) ± 4
(D) ± 5
Solution:
(C) ± 4

Question 2.
If the lines represented by kx² – 3xy + 6y² = 0 are perpendicular to each other then _________.
(A) k = 6
(B) k = -6
(C) k = 3
(D) k = -3
Solution:
(B) k = -6

Question 3.
Auxiliary equation of 2x² + 3xy – 9y² = 0 is _________.
(A) 2m² + 3m – 9 = 0
(B) 9m² – 3m – 2 = 0
(C) 2m² – 3m + 9 = 0
(D) -9m² – 3m + 2 = 0
Solution:
(B) 9m² – 3m – 2 = 0

Question 4.
The difference between the slopes of the lines represented by 3x² – 4xy + y² = 0 is _________.
(A) 2
(B) 1
(C) 3
(D) 4
Solution:
(A) 2

Question 5.
If the two lines ax² +2hxy+ by² = 0 make angles α and β with X-axis, then tan (α + β) = _____.
(A) \(\frac{h}{a+b}\)
(B) \(\frac{h}{a-b}\)
(C) \(\frac{2 h}{a+b}\)
(D) \(\frac{2 h}{a-b}\)
Solution:
(D) \(\frac{2 h}{a-b}\)

Question 6.
If the slope of one of the two lines \(\frac{x^{2}}{a}+\frac{2 x y}{h}+\frac{y^{2}}{b}\) = 0 is twice that of the other, then ab:h² = ___.
(A) 1 : 2
(B) 2 : 1
(C) 8 : 9
(D) 9 : 8
Solution:
(D) 9 : 8

Question 7.
The joint equation of the lines through the origin and perpendicular to the pair of lines 3x² + 4xy – 5y² = 0 is _________.
(A) 5x² + 4xy – 3y² = 0
(B) 3x² + 4xy – 5y² = 0
(C) 3x² – 4xy + 5y² = 0
(D) 5x² + 4xy + 3y² = 0
Solution:
(A) 5x² + 4xy – 3y² = 0

Question 8.
If acute angle between lines ax² + 2hxy + by² = 0 is, \(\frac{\pi}{4}\) then 4h² = _________.
(A) a² + 4ab + b²
(B) a² + 6ab + b²
(C) (a + 2b)(a + 3b)
(D) (a – 2b)(2a + b)
Solution:
(B) a² + 6ab + b²

Question 9.
If the equation 3x² – 8xy + qy² + 2x + 14y + p = 1 represents a pair of perpendicular lines then
the values of p and q are respectively _________.
(A) -3 and -7
(B) -7 and -3
(C) 3 and 7
(D) -7 and 3
Solution:
(B) -7 and -3

Question 10.
The area of triangle formed by the lines x² + 4xy + y² = 0 and x – y – 4 = 0 is _________.
(A) \(\frac{4}{\sqrt{3}}\) Sq. units
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
(C) \(\frac{16}{\sqrt{3}}\) Sq. units
(D)\(\frac{15}{\sqrt{3}}\) Sq. units
Solution:
(B) \(\frac{8}{\sqrt{3}}\) Sq. units
[Hint : Area = \(\frac{p^{2}}{\sqrt{3}}\), where p is the length of perpendicular from the origin to x – y – 4 = 0]

Question 11.
The combined equation of the co-ordinate axes is _________.
(A) x + y = 0
(B) x y = k
(C) xy = 0
(D) x – y = k
Solution:
(C) xy = 0

Question 12.
If h² = ab, then slope of lines ax² + 2hxy + by² = 0 are in the ratio _________.
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 1 : 1
Solution:
(D) 1 : 1
[Hint: If h² = ab, then lines are coincident. Therefore slopes of the lines are equal.]

Question 13.
If slope of one of the lines ax² + 2hxy + by² = 0 is 5 times the slope of the other, then 5h² = _________.
(A) ab
(B) 2 ab
(C) 7 ab
(D) 9 ab
Solution:
(D) 9 ab

Question 14.
If distance between lines (x – 2y)² + k(x – 2y) = 0 is 3 units, then k =
(A) ± 3
(B) ± 5\(\sqrt {5}\)
(C) 0
(D) ± 3\(\sqrt {5}\)
Solution:
(D) ± 3\(\sqrt {5}\)
[Hint: (x – 2y)² + k(x – 2y) = 0
∴ (x – 2y)(x – 2y + k) = 0
∴ equations of the lines are x – 2y = 0 and x – 2y + k = 0 which are parallel to each other.
∴ \(\left|\frac{k-0}{\sqrt{1+4}}\right|\) = 3
∴ k = ± 3\(\sqrt {5}\)

II. Solve the following.
Question 1.
Find the joint equation of lines:
(i) x – y = 0 and x + y = 0
Solution:
The joint equation of the lines x – y = 0 and
x + y = 0 is
(x – y)(x + y) = 0
∴ x² – y² = 0.

(ii) x + y – 3 = 0 and 2x + y – 1 = 0
Solution:
The joint equation of the lines x + y – 3 = 0 and 2x + y – 1 = 0 is
(x + y – 3)(2x + y – 1) = 0
∴ 2x² + xy – x + 2xy + y² – y – 6x – 3y + 3 = 0
∴ 2x² + 3xy + y² – 7x – 4y + 3 = 0.

(iii) Passing through the origin and having slopes 2 and 3.
Solution:
We know that the equation of the line passing through the origin and having slope m is y = mx. Equations of the lines passing through the origin and having slopes 2 and 3 are y = 2x and y = 3x respectively.
i.e. their equations are
2x – y = 0 and 3x – y = 0 respectively.
∴ their joint equation is (2x – y)(3x – y) = 0
∴ 6x² – 2xy – 3xy + y² = 0
∴ 6x² – 5xy + y² = 0.

(iv) Passing through the origin and having inclinations 60° and 120°.
Solution:
Slope of the line having inclination θ is tan θ .
Inclinations of the given lines are 60° and 120°
∴ their slopes are m₁ = tan60° = \(\sqrt {3}\) and
m₂ = tan 120° = tan (180° – 60°)
= -tan 60° = –\(\sqrt {3}\)
Since the lines pass through the origin, their equa-tions are
y = \(\sqrt {3}\)x and y= –\(\sqrt {3}\)x
i.e., \(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0
∴ the joint equation of these lines is
(\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ 3x² – y² = 0.

(v) Passing through (1, 2) amd parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0
∴ the equations of the lines passing through (1, 2) and parallel to the coordinate axes are x = 1 and y =1
i.e. x – 1 = 0 and y – 2 0
∴ their combined equation is
(x – 1)(y – 2) = 0
∴ x(y – 2) – 1(y – 2) = 0
∴ xy – 2x – y + 2 = 0

(vi) Passing through (3, 2) and parallel to the line x = 2 and y = 3.
Solution:
Equations of the lines passing through (3, 2) and parallel to the lines x = 2 and y = 3 are x = 3 and y = 2.
i.e. x – 3 = 0 and y – 2 = 0
∴ their joint equation is
(x – 3)(y – 2) = 0
∴ xy – 2x – 3y + 6 = 0.

(vii) Passing through (-1, 2) and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0.
Solution:
Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 respectively.
Slopes of the lines x + 2y + 3 = 0 and 3x – 4y – 5 = 0 are \(-\frac{1}{2}\) and \(-\frac{3}{-4}=\frac{3}{4}\) respectively.
∴ slopes of the lines L₁and L₂ are 2 and \(\frac{-4}{3}\) respectively.
Since the lines L₁ and L₂ pass through the point (-1, 2), their equations are
∴ (y – y₁) = m(x – x₁)
∴ (y – 2) = 2(x + 1)
⇒ y – 1 = 2x + 2
⇒ 2x – y + 4 = 0 and
∴ (y – 2) = \(\left(\frac{-4}{3}\right)\)(x + 1)
⇒ 3y – 6 = (-4)(x + 1)
⇒ 3y – 6 = -4x + 4
⇒ 4x + 3y – 6 + 4 = 0
⇒ 4x + 3y – 2 = 0
their combined equation is
∴ (2x – y + 4)(4x + 3y – 2) = 0
∴ 8x² + 6xy – 4x – 4xy – 3y² + 2y + 16x + 12y – 8 = 0
∴ 8x² + 2xy + 12x – 3y² + 14y – 8 = 0

(viii) Passing through the origin and having slopes 1 + \(\sqrt {3}\) and 1 – \(\sqrt {3}\)
Solution:
Let l₁ and l₂ be the two lines. Slopes of l₁ is 1 + \(\sqrt {3}\) and that of l₂ is 1 – \(\sqrt {3}\)
Therefore the equation of a line (l₁) passing through the origin and having slope is
y = (1 + \(\sqrt {3}\))x
∴ (1 + \(\sqrt {3}\))x – y = 0 ..(1)
Similarly, the equation of the line (l₂) passing through the origin and having slope is
y = (1 – \(\sqrt {3}\))x
∴ (1 – \(\sqrt {3}\))x – y = 0 …(2)
From (1) and (2) the required combined equation is

∴ (1 – 3)x² – 2xy + y² = 0
∴ -2x² – 2xy + y² = 0
∴ 2x² + 2xy – y² = 0
This is the required combined equation.

(ix) Which are at a distance of 9 units from the Y – axis.
Solution:
Equations of the lines, which are parallel to the Y-axis and at a distance of 9 units from it, are x = 9 and x = -9
i.e. x – 9 = 0 and x + 9 = 0

∴ their combined equation is
(x – 9)(x + 9) = 0
∴ x² – 81 = 0.

(x) Passing through the point (3, 2), one of which is parallel to the line x – 2y = 2 and other is perpendicular to the line y = 3.
Solution:
Let L₁ be the line passes through (3, 2) and parallel to the line x – 2y = 2 whose slope is \(\frac{-1}{-2}=\frac{1}{2}\)
∴ slope of the line L₁ is \(\frac{1}{2}\).
∴ equation of the line L₁ is
y – 2 = \(\frac{1}{2}\)(x – 3)
∴ 2y – 4 = x – 3 ∴ x – 2y + 1 = 0
Let L₂ be the line passes through (3, 2) and perpendicular to the line y = 3.
∴ equation of the line L₂ is of the form x = a.
Since L₂ passes through (3, 2), 3 = a
∴ equation of the line L₂ is x = 3, i.e. x – 3 = 0
Hence, the equations of the required lines are
x – 2y + 1 = 0 and x – 3 = 0
∴ their joint equation is
(x – 2y + 1)(x – 3) = 0
∴ x² – 2xy + x – 3x + 6y – 3 = 0
∴ x² – 2xy – 2x + 6y – 3 = 0.

(xi) Passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18.
Solution:
Let L₁ and L₂ be the lines passing through the origin and perpendicular to the lines x + 2y = 19 and 3x + y = 18 respectively.
Slopes of the lines x + 2y = 19 and 3x + y = 18 are \(-\frac{1}{2}\) and \(-\frac{3}{1}\) = -3 respectively.
Since the lines L₁ and L₂ pass through the origin, their equations are
y = 2x and y = \(\frac{1}{3}\)x
i.e. 2x – y = 0 and x – 3y = 0
∴ their combined equation is
(2x – y)(x – 3y) = 0
∴ 2x² – 6xy – xy + 3y² = 0
∴ 2x² – 7xy + 3y² = 0.

Question 2.
Show that each of the following equation represents a pair of lines.
(i) x² + 2xy – y² = 0
Solution:
Comparing the equation x² + 2xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 2, i.e. h = 1 and b = -1
∴ h² – ab = (1)² – 1(-1) = 1 + 1=2 > 0
Since the equation x² + 2xy – y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(ii) 4x² + 4xy + y² = 0
Solution:
Comparing the equation 4x² + 4xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 4, 2h = 4, i.e. h = 2 and b = 1
∴ h² – ab = (2)² – 4(1) = 4 – 4 = 0
Since the equation 4x² + 4xy + y² = 0 is a homogeneous equation of second degree and h² – ab = 0, the given equation represents a pair of lines which are real and coincident.

(iii) x² – y² = 0
Solution:
Comparing the equation x² – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 0, i.e. h = 0 and b = -1
∴ h² – ab = (0)² – 1(-1) = 0 + 1 = 1 > 0
Since the equation x² – y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(iv) x² + 7xy – 2y² = 0
Solution:
Comparing the equation x² + 7xy – 2y² = 0
a = 1, 2h = 7 i.e., h = \(\frac{7}{2}\) and b = -2
∴ h² – ab = \(\left(\frac{7}{2}\right)^{2}\) – 1(-2)
= \(\frac{49}{4}\) + 2
= \(\frac{57}{4}\) i.e. 14.25 = 14 > 0
Since the equation x² + 7xy – 2y² = 0 is a homogeneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

(v) x² – 2\(\sqrt {3}\) xy – y² = 0
Solution:
Comparing the equation x² – 2\(\sqrt {3}\) xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h= -2\(\sqrt {3}\), i.e. h = –\(\sqrt {3}\) and b = 1
∴ h² – ab = (-\(\sqrt {3}\))² – 1(1) = 3 – 1 = 2 > 0
Since the equation x² – 2\(\sqrt {3}\)xy – y² = 0 is a homo¬geneous equation of second degree and h² – ab > 0, the given equation represents a pair of lines which are real and distinct.

Question 3.
Find the separate equations of lines represented by the following equations:
(i) 6x² – 5xy – 6y² = 0
Solution:
6x² – 5xy – 6y² = 0
∴ 6x² – 9xy + 4xy – 6y² = 0
∴ 3x(2x – 3y) + 2y(2x – 3y) = 0
∴ (2x – 3y)(3x + 2y) = 0
∴ the separate equations of the lines are
2x – 3y = 0 and 3x + 2y = 0.

(ii) x² – 4y² = 0
Solution:
x² – 4y² = 0
∴ x² – (2y)² = 0
∴(x – 2y)(x + 2y) = 0
∴ the separate equations of the lines are
x – 2y = 0 and x + 2y = 0.

(iii) 3x² – y² = 0
Solution:
3x² – y² = 0
∴ (\(\sqrt {3}\) x)² – y² = 0
∴ (\(\sqrt {3}\)x – y)(\(\sqrt {3}\)x + y) = 0
∴ the separate equations of the lines are
\(\sqrt {3}\)x – y = 0 and \(\sqrt {3}\)x + y = 0.

(iv) 2x² + 2xy – y² = 0
Solution:
2x² + 2xy – y² = 0
∴ The auxiliary equation is -m² + 2m + 2 = 0
∴ m² – 2m – 2 = 0

m₁ = 1 + \(\sqrt {3}\) and m₂ = 1 – \(\sqrt {3}\) are the slopes of the lines.
∴ their separate equations are
y = m₁x and y = m₂x
i.e. y = (1 + \(\sqrt {3}\))x and y = (1 – \(\sqrt {3}\))x
i.e. (\(\sqrt {3}\) + 1)x – y = 0 and (\(\sqrt {3}\) – 1)x + y = 0.

Question 4.
Find the joint equation of the pair of lines through the origin and perpendicular to the lines
given by :
(i) x² + 4xy – 5y² = 0
Solution:
Comparing the equation x² + 4xy – 5y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 4, b= -5
Let m₁ and m₂ be the slopes of the lines represented by x² + 4xy – 5y² = 0.

Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{4}{5}\)xy – \(\frac{1}{5}\)y² = 0 …[By (1)]
∴ 5x² + 4xy – y² = 0

(ii) 2x² – 3xy – 9y² = 0
Solution:
Comparing the equation 2x² – 3xy – 9y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 2, 2h = -3, b = -9
Let m₁ and m₂ be the slopes of the lines represented by 2x² – 3xy – 9y² = 0
∴ m₁ + m₂ =\(\frac{-2 h}{b}=-\frac{3}{9}\) and m₁m₂ = \(\frac{a}{b}=-\frac{2}{9}\) …(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\left(-\frac{3}{9}\right)\)xy + \(\left(-\frac{2}{9}\right)\)y² = 0 …[By (1)]
∴ 9x² – 3xy – 2y² = 0

(iii) x² + xy – y² = 0
Solution:
Comparing the equation x²+ xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by x² + xy – y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-1}{-1}\) and m₁m₂ = \(\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{-1}\) = -1 ..(1)
Now, required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂) + m₁m₂y² = 0
∴ x² + 1xy + (-1)y² = 0 …[By (1)]
∴ x² + xy – y² = 0

Question 5.
Find k if
(i) The sum of the slopes of the lines given by 3x² + kxy – y² = 0 is zero.
Solution:
Comparing the equation 3x² + kxy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 3, 2h = k, b = -1
Let m₁ and m₂ be the slopes of the lines represented by 3x² + kxy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-k}{-1}\) = k
Now, m₁ + m₂ = 0 … (Given)
∴ k = 0.

(ii) The sum of slopes of the lines given by 2x² + kxy – 3y² = 0 is equal to their product.
Question is modified.
The sum of slopes of the lines given by x² + kxy – 3y² = 0 is equal to their product.
Solution:
Comparing the equation x² + kxy – 3y² = 0, with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = k, b = -3
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}=\frac{-k}{-3}=\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}=\frac{1}{-3}=\frac{-1}{3}\)
Now, m₁ + m₂ = m₁m₂ … (Given)
∴ \(\frac{k}{3}=\frac{-1}{3}\)
∴ k = -1.

(iii) The slope of one of the lines given by 3x² – 4xy + ky² = 0 is 1.
Solution:
The auxiliary equation of the lines given by 3x² – 4xy + ky² = 0 is km² – 4m + 3 = 0.
Given, slope of one of the lines is 1.
∴ m = 1 is the root of the auxiliary equation km² – 4m + 3 = 0.
∴ k(1)² – 4(1) + 3 = 0
∴ k – 4 + 3 = 0
∴ k = 1.

(iv) One of the lines given by 3x² – kxy + 5y² = 0 is perpendicular to the 5x + 3y = 0.
Solution:
The auxiliary equation of the lines represented by 3x² – kxy + 5y² = 0 is 5m² – km + 3 = 0.
Now, one line is perpendicular to the line 5x + 3y = 0, whose slope is \(-\frac{5}{3}\).
∴ slope of that line = m = \(\frac{3}{5}\)
∴ m = \(\frac{3}{5}\) is the root of the auxiliary equation 5
5m² – km + 3 = 0.
∴ 5\(\left(\frac{3}{5}\right)^{2}\) – k\(\left(\frac{3}{5}\right)\) + 3 = 0
∴ \(\frac{9}{5}-\frac{3 k}{5}\) + 3 = 0
∴ 9 – 3k + 15 = 0
∴ 3k = 24
∴ k = 8.

(v) The slope of one of the lines given by 3x² + 4xy + ky² = 0 is three times the other.
Solution:
3x² + 4xy + ky² = 0
∴ divide by x²

∴ y = mx
∴ \(\frac{\mathrm{y}}{\mathrm{x}}\) = m
put \(\frac{\mathrm{y}}{\mathrm{x}}\) = m in equation (1)
Comparing the equation km² + 4m + 3 = 0 with ax² + 2hxy+ by² = 0, we get,
a = k, 2h = 4, b = 3
m₁ = 3m₂ ..(given condition)
m₁ + m₂ = \(\frac{-2 h}{k}=-\frac{4}{k}\)
m₁m₂ = \(\frac{a}{b}=\frac{3}{k}\)
m₁ + m₂ = \(-\frac{4}{\mathrm{k}}\)
4m₂ = \(-\frac{4}{\mathrm{k}}\) …(m₁ = 3m₂)
m₂ = \(-\frac{1}{\mathrm{k}}\)
m₁m₂ = \(\frac{3}{k}\)
\(3 \mathrm{~m}_{2}^{2}=\frac{3}{\mathrm{k}}\) …(m₁ = 3m₂)
\(3\left(-\frac{1}{\mathrm{k}}\right)^{2}=\frac{3}{\mathrm{k}}\) …(m₂ = \(-\frac{1}{k}\))
\(\frac{1}{k^{2}}=\frac{1}{k}\)
k² = k
k = 1 or k = 0

(vi) The slopes of lines given by kx² + 5xy + y² = 0 differ by 1.
Solution:
Comparing the equation kx² + 5xy +y² = 0 with ax² + 2hxy + by²
a = k, 2h = 5 i.e. h = \(\frac{5}{2}\)
m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{5}{1}\) = -5
and m₁m₂ = \(\frac{a}{b}=\frac{k}{1}\) = k
the slope of the line differ by (m₁ – m₂) = 1 …(1)
∴ (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂
(m₁ – m₂)² = (-5)² – 4(k)
(m₁ – m₂)² = 25 – 4k
1 = 25 – 4k ..[By (1)]
4k = 24
k = 6

(vii) One of the lines given by 6x² + kxy + y² = 0 is 2x + y = 0.
Solution:
The auxiliary equation of the lines represented by 6x² + kxy + y² = 0 is
m² + km + 6 = 0.
Since one of the line is 2x + y = 0 whose slope is m = -2.
∴ m = -2 is the root of the auxiliary equation m² + km + 6 = 0.
∴ (-2)² + k(-2) + 6 = 0
∴ 4 – 2k + 6 = 0
∴ 2k = 10 ∴ k = 5

Question 6.
Find the joint equation of the pair of lines which bisect angle between the lines given by x² + 3xy + 2y² = 0
Solution:
x² + 3xy + 2y² = 0
∴ x² + 2xy + xy + 2y² = 0
∴ x(x + 2y) + y(x + 2y) = 0
∴ (x + 2y)(x + y) = 0
∴ separate equations of the lines represented by x² + 3xy + 2y² = 0 are x + 2y = 0 and x + y = 0.
Let P (x, y) be any point on one of the angle bisector. Since the points on the angle bisectors are equidistant from both the lines,

the distance of P (x, y) from the line x + 2y = 0
= the distance of P(x, y) from the line x + y = 0

∴ 2(x + 2y)² = 5(x + y)²
∴ 2(x² + 4xy + 4y²) = 5(x² + 2xy + y²)
∴ 2x² + 8xy + 8y² = 5x² + 10xy + 5y²
∴ 3x² + 2xy – 3y² = 0.
This is the required joint equation of the lines which bisect the angles between the lines represented by x² + 3xy + 2y² = 0.

Question 7.
Find the joint equation of the pair of lies through the origin and making equilateral triangle with the line x = 3.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the line x = 3.
∴ OA and OB make an angle of 30° and 150° with the positive direction of X-axis
∴ slope of OA = tan 30° = 1/\(\sqrt {3}\)
∴ equation of the line OA is y = \(\frac{1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = x ∴ x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= – tan 30°= -1/\(\sqrt {3}\)
∴ equation of the line OB is y = \(\frac{-1}{\sqrt{3}}\)x
∴ \(\sqrt {3}\)y = -x ∴ x + \(\sqrt {3}\)y = 0
∴ required combined equation of the lines is
(x – \(\sqrt {3}\)y) (x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Question 8.
Show that the lines x² – 4xy + y² = 0 and x + y = 10 contain the sides of an equilateral triangle. Find the area of the triangle.
Solution:
We find the joint equation of the pair of lines OA and OB through origin, each making an angle of 60° with x + y = 10 whose slope is -1.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
Also, tan 60° = \(\left|\frac{m-(-1)}{1+m(-1)}\right|\)
∴ \(\sqrt {3}\) = \(\left|\frac{m+1}{1-m}\right|\)
Squaring both sides, we get,
3 = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ 3(1 – 2m + m²) = m² + 2m + 1
∴ 3 – 6m + 3m² = m² + 2m + 1
∴ 2m² – 8m + 2 = 0
∴ m² – 4m + 1 = 0
∴ \(\left(\frac{y}{x}\right)^{2}\) – 4\(\left(\frac{y}{x}\right)\) + 1 = 0 …[By (1)]
∴ y² – 4xy + x² = 0
∴ x² – 4xy + y\left(\frac{y}{x}\right) = 0 is the joint equation of the two lines through the origin each making an angle of 60° with x + y = 10
∴ x² – 4xy + y² = 0 and x + y = 10 form a triangle OAB which is equilateral.
Let seg OM ⊥^r line AB whose question is x + y = 10

Question 9.
If the slope of one of the lines represented by ax² + 2hxy + by² = 0 is three times the other then prove that 3h² = 4ab.
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
We are given that m₂ = 3m₁
∴ m₁ + 3m₁ = \(-\frac{2 h}{b}\) 4m₁ = \(-\frac{2 h}{b}\)
∴ m₁ = \(-\frac{h}{2 b}\) …(1)
Also, m₁(3m₁) = \(\frac{a}{b}\) ∴ 3m₁² = \(\frac{a}{b}\)
∴ 3\(\left(-\frac{h}{2 b}\right)^{2}\) = \(\frac{a}{b}\) ….[By (1)]
∴ \(\frac{3 h^{2}}{4 b^{2}}=\frac{a}{b}\)
∴ 3h² = 4ab, as b ≠0.

Question 10.
Find the combined equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.
Solution:
Comparing the equation 5x² + 6xy – y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 6, b = -1
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 6xy – y² = 0.

The separate equations of the lines are
y = m₁x and y = m₂x, where m1 ≠ m2
i.e. m₁x – y = 0 and m₁x – y = 0.
Let P (x, y) be any point on one of the bisector of the angles between the lines.
∴ the distance of P from the line m₁x – y = 0 is equal to the distance of P from the line m₂x – y = 0.

∴ (m₂² + 1)(m₁x – y)² = (m₁² + 1)(m₂x – y)²
∴ (m₂² + 1)(m₁²x² – 2m₁xy + y²) = (m₁² + 1)(m₂²x² – 2m₂xy + y²)
∴ m₁²m₂²x² – 2m₁m₁²y²xy + m₂²y² + m₁²x² – 2m₁²xy + y²
= m₁²m₂²x² – 2m₁²m₂xy + m₁²y² + m₂²x² – 2m₂xy + y²
∴ (m₁² – m₂²)x² + 2m₁m₂(m₁ – m₂)xy – 2(m₁ – m₂)xy – (m₁² – m₂²)y² = 0
Dividing throughout by m₁ – m₂ (≠0), we get,
(m₁ + m₂)x² + 2m₁m₂xy – 2xy – (m₁ + m₂)y² = 0
∴ 6x² – 10xy – 2xy – 6y² = 0 …[By (1)]
∴ 6x² – 12xy – 6y² = 0
∴ x² – 2xy – y² = 0
This is the joint equation of the bisectors of the angles between the lines represented by 5x² + 6xy – y² = 0.

Question 11.
Find a, if the sum of the slopes of the lines represented by ax² + 8xy + 5y² = 0 is twice their product.
Solution :
Comparing the equation ax² + 8xy + 5y² = 0 with ax² + 2hxy + by² = 0,
we get, a = a, 2h = 8, b = 5
Let m₁ and m₂ be the slopes of the lines represented by ax² + 8xy + 5y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{8}{5}\)
and m₁m₂ = \(\frac{a}{b}=\frac{a}{5}\)
Now, (m₁ + m₂) = 2(m₁m₂)
\(-\frac{8}{5}\) = \(2\left(\frac{a}{5}\right)\)
a = -4

Question 12.
If the line 4x – 5y = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0, then show that 25a + 40h +16b = 0.
Solution :
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0
Given that 4x – 5y = 0 is one of the lines represented by ax² + 2hxy + by² = 0.
The slope of the line 4x – 5y = 0 is \(\frac{-4}{-5}=\frac{4}{5}\)
∴ m = \(\frac{4}{5}\) is a root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(\frac{4}{5}\right)^{2}\) + 2h\(\left(\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}+\frac{8 h}{5}\) + a = 0
∴ 16b + 40h + 25a = 0 i.e.
∴ 25a + 40h + 16b = 0

Question 13.
Show that the following equations represent a pair of lines. Find the acute angle between them :
(i) 9x² – 6xy + y² + 18x – 6y + 8 = 0
Solution:
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 9, h = -3, b = 1, g = 9, f = -3 and c = 8.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
9 & -3 & 9 \\
-3 & 1 & -3 \\
9 & -3 & 8
\end{array}\right|\)
= 9(8 – 9) + 3(-24 + 27) + 9(9 – 9)
= 9(-1) + 3(3) + 9(0)
= -9 + 9 + 0 = 0
and h² – ab = (-3)² – 9(1) = 9 – 9 = 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.

∴ tan θ = tan0°
∴ θ = 0°.

(ii) 2x² + xy – y² + x + 4y – 3 = 0
Solution:
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy+ c = 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2 and c = -3

= -2 + 1 + 1
= -2 + 2= 0
∴ the given equation represents a pair of lines.
Let θ be the acute angle between the lines.

∴ tan θ = tan 3
∴ θ = tan^-1(3)

(iii) (x – 3)² + (x – 3)(y – 4) – 2(y – 4)² = 0.
Solution :
Put x – 3 = X and y – 4 = Y in the given equation, we get,
X² + XY – 2Y² = 0
Comparing this equation with ax² + 2hxy + by² = 0, we get,
a = 1, h = \(\frac{1}{2}\), b = -2
This is the homogeneous equation of second degreeand h² – ab = \(\left(\frac{1}{2}\right)^{2}\) – 1(-2)
= \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
Hence, it represents a pair of lines passing through the new origin (3, 4).
Let θ be the acute angle between the lines.

∴ tanθ = 3 ∴ θ = tan^-1(3)

Question 14.
Find the combined equation of pair of lines through the origin each of which makes angle of 60° with the Y-axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Question 15.
If lines representedby ax² + 2hxy + by² = 0 make angles of equal measures with the co-ordinate
axes then show that a = ± b.
OR
Show that, one of the lines represented by ax² + 2hxy + by² = 0 will make an angle of the same measure with the X-axis as the other makes with the Y-axis, if a = ± b.
Solution:

Let OA and OB be the two lines through the origin represented by ax² + 2hxy + by² = 0.
Since these lines make angles of equal measure with the coordinate axes, they make angles ∝ and \(\frac{\pi}{2}\) – ∝ with the positive direction of X-axis or ∝ and \(\frac{\pi}{2}\) + ∝ with thepositive direction of X-axis.
∴ slope of the line OA = m₁ = tan ∝
and slope of the line OB = m₂
= tan(\(\frac{\pi}{2}\) – ∝) or tan(\(\frac{\pi}{2}\) + ∝)
i.e. m₂ = cot ∝ or m₂ = -cot ∝
∴ m₁m₂ – tan ∝ x cot ∝ = 1
OR m₁m₂ = tan ∝ (-cot ∝) = -1
i.e. m₁m₂ = ± 1
But m₁m₂ = \(\frac{a}{b}\)
∴ \(\frac{a}{b}\)= ±1 ∴ a = ±b
This is the required condition.

Question 16.
Show that the combined equation of a pair of lines through the origin and each making an angle of ∝ with the line x + y = 0 is x² + 2(sec 2∝) xy + y² = 0.
Solution:
Let OA and OB be the required lines.
Let OA (or OB) has slope m.
∴ its equation is y = mx … (1)
It makes an angle ∝ with x + y = 0 whose slope is -1. m +1
∴ tan ∝ = \(\left|\frac{m+1}{1+m(-1)}\right|\)
Squaring both sides, we get,
tan²∝ = \(\frac{(m+1)^{2}}{(1-m)^{2}}\)
∴ tan²∝(1 – 2m + m²) = m² + 2m + 1
∴ tan²∝ – 2m tan²∝ + m²tan²∝ = m² + 2m + 1
∴ (tan²∝ – 1)m² – 2(1 + tan²∝)m + (tan²∝ – 1) = 0

∴ y² + 2xysec2∝ + x² = 0
∴ x² + 2(sec2∝)xy + y² = 0 is the required equation.

Question 17.
Show that the line 3x + 4y+ 5 = 0 and the lines (3x + 4y)² – 3(4x – 3y)² =0 form an equilateral triangle.
Solution:
The slope of the line 3x + 4y + 5 = 0 is \(\frac{-3}{4}\)
Let m be the slope of one of the line making an angle of 60° with the line 3x + 4y + 5 = 0. The angle between the lines having slope m and m₁ is 60°.

On squaring both sides, we get,
3 = \(\frac{(4 m+3)^{2}}{(4-3 m)^{2}}\)
∴ 3 (4 – 3m)² = (4m + 3)²
∴ 3(16 – 24m + 9m²) = 16m² + 24m + 9
∴ 48 – 72m + 27m² = 16m² + 24m + 9
∴ 11m² – 96m + 39 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
11\(\left(\frac{y}{x}\right)^{2}\) – 96\(\left(\frac{y}{x}\right)\) + 39 = 0
∴ \(\frac{11 y^{2}}{x^{2}}-\frac{96 y}{x}\) + 39 = 0
∴ 11y² – 96xy + 39x² = 0
∴ 39x² – 96xy + 11y² = 0.
∴ 39x² – 96xy + 11y² = 0 is the joint equation of the two lines through the origin each making an angle of 60° with the line 3x + 4y + 5 = 0.
The equation 39x² – 96xy + 11y² = 0 can be written as :
-39x² + 96xy – 11y² = 0
i.e., (9x² – 48x²) + (24xy + 72xy) + (16y² – 27y²) = 0
i.e. (9x² + 24xy + 16y²) – (48x² – 72xy + 27y²) = 0
i.e. (9x² + 24xy + 16y²) – 3(16x² – 24xy + 9y²) = 0
i.e. (3x + 4y)² – 3(4x – 3y)² = 0
Hence, the line 3x + 4y + 5 = 0 and the lines
(3x + 4y)² – 3(4x – 3y)² form the sides of an equilateral triangle.

Question 18.
Show that lines x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form an equilateral triangle. Find its area and perimeter.
Solution:
x² – 4xy + y² = 0 and x + y = \(\sqrt {6}\) form a triangle OAB which is equilateral.
Let OM be the perpendicular from the origin O to AB whose equation is x + y = \(\sqrt {6}\)

In right angled triangle OAM,
sin 60° = \(\frac{\mathrm{OM}}{\mathrm{OA}}\) ∴ \(\frac{\sqrt{3}}{2}\) = \(\frac{\sqrt{3}}{\mathrm{OA}}\)
∴ OA = 2
∴ length of the each side of the equilateral triangle OAB = 2 units.
∴ perimeter of ∆ OAB = 3 × length of each side
= 3 × 2 = 6 units.

Question 19.
If the slope of one of the lines given by ax² + 2hxy + by² = 0 is square of the other then show that a²b + ab² + 8h³ = 6abh.
Solution:
Let m be the slope of one of the lines given by ax² + 2hxy + by² = 0.
Then the other line has slope m²

Multiplying by b³, we get,
-8h³ = ab² + a²b – 6abh
∴ a²b + ab² + 8h³ = 6abh
This is the required condition.

Question 20.
Prove that the product of lengths of perpendiculars drawn from P (x₁, y₁) to the lines repersented by ax² + 2hxy + by² = 0 is \(\left|\frac{a x_{1}^{2}+2 h x_{1} y_{1}+b y_{1}^{2}}{\sqrt{(a-b)^{2}+4 h^{2}}}\right|\)
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.
∴ m¹ + m² = \(-\frac{2 h}{b}\) and m¹m² = \(\frac{a}{b}\) …(1)
The separate equations of the lines represented by
ax² + 2hxy + by² = 0 are
y = m₁x and y = m₂x
i.e. m₁x – y = 0 and m₂x – y = 0
Length of perpendicular from P(x₁, ₁) on

Question 21.
Show that the difference between the slopes of lines given by (tan²θ + cos²θ )x² – 2xytanθ + (sin²θ )y² = 0 is two.
Solution:
Comparing the equation (tan²θ + cos²θ)x² – 2xy tan θ + (sin²θ) y² = 0 with ax² + 2hxy + by² = 0, we get,
a = tan²θ + cos²θ, 2h = -2 tan θ and b = sin2θ
Let m₁ and m₂ be the slopes of the lines represented by the given equation.

Question 22.
Find the condition that the equation ay² + bxy + ex + dy = 0 may represent a pair of lines.
Solution:
Comparing the equation
ay² + bxy + ex + dy = 0 with
Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0, we get,
A = 0, H = \(\frac{b}{2}\), B = a,G = \(\frac{e}{2}\), F = \(\frac{d}{2}\), C = 0
The given equation represents a pair of lines,

i.e. if bed – ae² = 0
i.e. if e(bd – ae) = 0
i.e. e = 0 or bd – ae = 0
i.e. e = 0 or bd = ae
This is the required condition.

Question 23.
If the lines given by ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1 then show that (3a + b)(a + 3b) = 4h².
Solution:
Since the lines ax² + 2hxy + by² = 0 form an equilateral triangle with the line lx + my = 1, the angle between the lines ax² + 2hxy + by² = 0 is 60°.

∴ 3(a + b)² = 4(h² – ab)
∴ 3(a² + 2ab + b²) = 4h² – 4ab
∴ 3a² + 6ab + 3b² + 4ab = 4h²
∴ 3a² + 10ab + 3b² = 4h²
∴ 3a² + 9ab + ab + 3b² = 4h²
∴ 3a(a + 3b) + b(a + 3b) = 4h²
∴ (3a + b)(a + 3b) = 4h²
This is the required condition.

Question 24.
If line x + 2 = 0 coincides with one of the lines represented by the equation x² + 2xy + 4y + k = 0 then show that k = -4.
Solution:
One of the lines represented by
x² + 2xy + 4y + k = 0 … (1)
is x + 2 = 0.
Let the other line represented by (1) be ax + by + c = 0.
∴ their combined equation is (x + 2)(ax + by + c) = 0
∴ ax² + bxy + cx + 2ax + 2by + 2c = 0
∴ ax² + bxy + (2a + c)x + 2by + 2c — 0 … (2)
As the equations (1) and (2) are the combined equations of the same two lines, they are identical.
∴ by comparing their corresponding coefficients, we get,

∴ 1 = \(\frac{-4}{k}\)
∴ k = -4.

Question 25.
Prove that the combined equation of the pair of lines passing through the origin and perpendicular to the lines represented by ax² + 2hxy + by² = 0 is bx² – 2hxy + ay² = 0
Solution:
Let m₁ and m₂ be the slopes of the lines represented by ax² + 2hxy + by² = 0.

Now, required lines are perpendicular to these lines.
∴ their slopes are and \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(-\frac{1}{m_{1}}\)x and y = \(-\frac{1}{m_{2}}\)x
i.e. m₁y= -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y)(x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x²\(\frac{-2 h}{b}\)x + \(\frac{a}{b}\)y² = 0
∴ bx² – 2hxy + ay² = 0.

Question 26.
If equation ax² – y² + 2y + c = 1 represents a pair of perpendicular lines then find a and c.
Solution:
The given equation represents a pair of lines perpendicular to each other.
∴ coefficient of x² + coefficient of y² = 0
∴ a – 1 = 0 ∴ a = 1
With this value of a, the given equation is
x² – y² + 2y + c – 1 = 0
Comparing this equation with
Ax² + 2Hxy + By² + 2Gx + 2Fy + C = 0, we get,
A = 1, H = 0, B = -1, G = 0, F = 1, C = c – 1
Since the given equation represents a pair of lines,
D = \(\left|\begin{array}{ccc}
A & H & G \\
H & B & F \\
G & F & C
\end{array}\right|\) = 0
∴ \(\left|\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 1 \\
0 & 1 & c-1
\end{array}\right|\) = 0
∴ 1(-c + 1 – 1) – 0 + 0 = 0
∴ -c = 0
∴ c = 0.
Hence, a = 1, c = 0.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 2 Trigonometry – I Ex 2.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 2 Trigonometry – I Ex 2.1

Question 1.
Find the trigonometric functions of 0°, 30°, 45°, 60°, 150°, 180°, 210°, 300°, 330°, – 30°, – 45°, – 60°, – 90°, – 120°, – 225°, – 240°, – 270°, – 315°
Solution:
Angle of measure 0°:

Let m∠XOA = 0° = 0^c
Its terminal arm (ray OA) intersects the standard
unit circle in P(1, 0).
Hence,x = 1 and y = 0
sin 0° = y = 0,
cos 0° = x = 1,
tan 0° = \(\frac{y}{x}=\frac{0}{1}\) = 0
cot 0° = \(\frac{x}{y}=\frac{1}{0}\) which is not defined
sec 0° = \(\frac{1}{x}=\frac{1}{1}\) = 1
cot 0° = \(\frac{1}{y}=\frac{1}{0}\) which is not defined,

Angle of measure 30°:
Let m∠XOA = 30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y)
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1

Since point P lies in 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{\sqrt{3}}{2}\) and y = PM = \(\frac{1}{2}\)

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0, y > 0
∴ x = OM = \(\frac{1}{\sqrt{2}}\) and
y = PM = \(\frac{1}{\sqrt{2}}\)
∴ P = (\(\frac{1}{\sqrt{2}}\), \(\frac{1}{\sqrt{2}}\))

Angle of measure 60°:
Let m∠XOA = 60°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Angle of measure 150°:
Let m∠XOA = 150°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 180°:
Let m∠XOA = 180°
Its terminal arm (ray OA) intersects the standard unit circle at P(-1, 0).
∴ x = – 1 and y = 0
sin 180° =y = 0
cos 180° = x = -1

tan 180° = \(\frac{y}{x}\)
= \(\frac{0}{-1}\) = 0
Cosec 180° = \(\frac{1}{y}\)
= \(\frac{1}{0}\)
which is not defined.
sec 180°= \(\frac{1}{x}=\frac{1}{-1}\) = -1
cot 180° = \(\frac{x}{y}=\frac{-1}{0}\) , which is not defined.

Angle of measure 210°:
Let m∠XOA = 210°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 3rd quadrant, x < 0,y < 0
∴ x = -OM = \(\frac{-\sqrt{3}}{2}\) and y = -PM = \(\frac{-1}{2}\)
∴ P ≡( \(\frac{-\sqrt{3}}{2}, \frac{-1}{2}\) )

Angle of measure 300°:
Let m∠XOA = 300°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 1st quadrant, x > 0,y > 0
x = OM = \(\frac{1}{2}\) = and y = -PM = \(\frac{-\sqrt{3}}{2}\)
sin 300° = y = \(\frac{-\sqrt{3}}{2}\)
cos 300° = x = \(\frac{1}{2}\)
tan 300° = \(\frac{y}{x}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}\)

Angle of measure 330°:
Let m∠XOA = 330°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP= 1,

Since point P lies in the 4th quadrant, x > 0, y < 0

Angle of measure 30°
Let m∠XOA = -30°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60 — 90° triangle.
op = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

Angle of measure 45°:
Let m∠XOA = 45°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 4th quadrant x > 0, y < 0

[Note : Answer given in the textbook of sin (45°) = – 1/2. However, as per our calculation it is \(-\frac{1}{\sqrt{2}}\) ]

Angle of measure (-60°):
Let m∠XOA = -60°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 4’ quadrant,
x > 0, y < 0
x = OM =\(\frac{1}{2}\) and y = -PM = \(-\frac{\sqrt{3}}{2}\)

Angle of measure (-90°):
Let m∠XOA = -90°
It terminal arm (ray OA) intersects the standard unit circle at P(0, -1)
∴ x = 0 and y = -1
sin (-90°) = y = -1
cos (-90°) = s = 0

Angle of measure (-120°):
Let m∠XOA = – 120°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30° – 60° – 90° triangle.
OP = 1,

Since point P lies in the 3rd quadrant, x < 0, y < 0

Angle of measure (- 225°):
Let m∠XOA = – 225°
Its terminal arm (ray OA) intersects the standard unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Since point P lies in the 2nd quadrant, x < 0, y > 0

Angle of measure 2400):
Let m∠XOA = 240°
Its terminal arm (ray OA) intersects the standard
unit circle at P(x, y).
Draw seg PM perpendicular to the X-axis.
∴ ΔOMP is a 30°  – 60° –  900 triangle.

Since point P lies in the 2nd quadrant, x<0, y>0

Angle of measure (- 270°):
Let m∠XOA = – 270°
Its terminal arm (ray OA)
intersects the standard unit,
circle at P(0, 1).
∴ x = 0 and y = 1
sin (- 270°) = y = 1
cos (- 270°) = x = 0
tan(-270°)= \(\frac{y}{x}=\frac{1}{0}\)
which is not defined.

Angle of measure ( 315°):
Let m∠XOA 315°
Its terminal arm (ray OA) intersects the standard unit circle at P(x,y).
Draw seg PM perpendicular to the X-axis.
ΔOMP is a 45° – 45° – 90° triangle.
OP = 1,

Question 2.
State the signs of:
i. tan 380°
ii. cot 230°
iii 468°
Solution:
1. 380° = 360° + 20°
∴ 380° and 20° are co-terminal angles.
Since 0° < 20° <90°0,
20° lies in the l quadrant.
∴ 380° lies in the 1st quadrant,
∴ tan 380° is positive.

ii. Since, 180° <230° <270°
∴ 230° lies in the 3rd quadrant.
∴ cot 230° is positive.

iii. 468° = 360°+108°
∴ 468° and 108° are co-terminal angles.
Since 90° < 108° < 180°,
108° lies in the 2nd quadrant.
∴ 468° lies in the 2nd quadrant.
∴ sec 468° is negative.

Question 3.
State the signs of cos 4^c and cos 4°. Which of these two functions is greater?
Solution:
Since 0° < 4° < 90°, 4° lies in the first quadrant. ∴ cos4° >0 …(i)
Since 1^c = 57° nearly,
180° < 4^c < 270°
∴ 4^c lies in the third quadrant.
∴ cos 4^c < 0 ………(ii)
From (i) and (ii),
cos 4° is greater.

Question 4.
State the quadrant in which 6 lies if
i. sin θ < 0 and tan θ > 0
ii. cos θ < 0 and tan θ > 0
Solution:
i. sin θ < 0 sin θ is negative in 3rd and 4th quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

ii. cos θ < 0 cos θ is negative in 2nd and 3rd quadrants, tan 0 > 0
tan θ is positive in 1st and 3rd quadrants.
∴ θ lies in the 3rd quadrant.

Question 5.
Evaluate each of the following:
i. sin 30° + cos 45° + tan 180°
ii. cosec 45° + cot 45° + tan 0°
iii. sin 30° x cos 45° x lies tan 360°
Solution:
i. We know that,
sin30° = 1/2, cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 180° = 0
sin30° + cos 45° +tan 180°
= \(\frac{1}{2}+\frac{1}{\sqrt{2}}+0=\frac{\sqrt{2}+1}{2}\)

ii. We know that,
cosec 45° = \(\sqrt{2}\) , cot 45° = 1, tan 0° = 0
cosec 45° + cot 45° + tan 0°
= \(\sqrt{2}\) + 1 + 0 = \(\sqrt{2}\) + 1

iii. We know that,
sin 30° = \(\frac{1}{2}\), cos 45° = \(\frac{1}{\sqrt{2}}\) =, tan 360° = 0
sin 30° x cos 45° x tan 360°
= \(\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\) = 0

Question 6.
Find all trigonometric functions of angle in standard position whose terminal arm passes through point (3, – 4).
Solution:
Let θ be the measure of the angle in standard position whose terminal arm passes through P(3, -4).
∴ x = 3 and y = -4
r = OP

Question 7.
If cos θ = \(\frac{12}{13}, 0<\theta<\frac{\pi}{2}\) find the value of \(\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cos \theta}, \frac{1}{\tan ^{2} \theta}\)
Solution:
cos θ = \(\frac{12}{13}\)
We know that,
sin² θ = 1 – cos²θ

∴ sin θ = ± \(\frac{5}{13}\)
Since 0 < θ < \(\frac{\pi}{2}\) , θ lies in the 1st quadrant, ∴ sin θ > 0

Question 8.
Using tables evaluate the following:
i. 4 cot 45° – sec2 60° + sin 30°
ii.\(\cos ^{2} 0+\cos ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \frac{\pi}{2}\)
Solution:
i. We know that,
cot 45° = 1, sec 60° = 2, sin 30° = 1/2
4 cot 45° – sec² 60° + sin 30°
= 4(1) – (2)² + \(\frac{1}{2}\)
= 4 – 4 + \(\frac{1}{2}=\frac{1}{2}\)

ii. We know that,

Question 9.
Find the other trigonometric functions if
i. cot θ = \(-\frac{3}{5}\), and 180 < θ < 270
ii. Sec A = \(-\frac{25}{7}\) and A lies in the second quadrant.
iii cot x = \(\frac{3}{4}\), x lies in the third quadrant.
iv. tan x = \(\frac{-5}{12}\) x lies in the fourth quadrant.
Solution:
i. cot θ = \(-\frac{3}{5}\)
we know that,
sin²θ = 1 – cos²θ
= 1 – \(\left(-\frac{3}{5}\right)^{2}\)
= 1 – \(\frac{9}{25}=\frac{16}{25}\)
∴ sin θ = ± \(\frac{4}{5}\)
Since 180° < 0 < 270°,
θ lies in the 3rd quadrant.
∴ sin θ < 0

Since A lies in the 2nd quadrant,
tan A < 0

iii. Given, cot x = \(\frac{3}{4}\)
We know that,
cosec² x = 1 + cot² x
= 1 + \(\left(\frac{3}{4}\right)^{2}=1+\frac{9}{16}=\frac{25}{16}\)
∴ cosec x = ± \(\frac{5}{4}\)
Since x lies in the 3rd quadrant, cosec x < 0

iv. Given, tan x = \(-\frac{5}{12}\)
sec² x = 1 + tan²
= 1 + \(\left(-\frac{5}{12}\right)^{2}\)
= 1 + \(\frac{25}{144}=\frac{169}{144}\)
∴ sec x = ± \(\frac{13}{12}\)
Since x lies in the 4th quadrant,
sec x > 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 1 Angle and its Measurement Miscellaneous Exercise 1

I. Select the correct option from the given alternatives.

Question 1.
\(\left(\frac{22 \pi}{15}\right)^{c}x\) is equal to
(A) 246°
(B) 264°
(C) 224°
(D) 426°
Answer:
(B) 264°

Question 2.
156° is equal to

Answer:
(B)

Question 3.
A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 metres when it traces the angle of 12° at the centre, then the length of the rope is
(A) 70 m
(B) 55 m
(C) 40 m
(D) 35 m
Answer:
(A) 70 m

Question 4.
A pendulum 14 cm long oscillates through an angle of 12°, then the angle of the path described by its extremities is


Answer:
(D)

Question 5.
Angle between hands of a clock when it shows the time 9 :45 is
(A) (7.5)°
(B) (12.5)°
(C) (17.5)°
(D) (22.5)°
Answer:
(D) (22.5)°

Question 6.
20 metres of wire is available for fencing off a flower-bed in the form of a circular sector of radius 5 metres, then .the maximum area (in sq. m.) of the flower-bed is
(A) 15
(B) 20
(C) 25
(D) 30
Answer:
(C) 25
r + r + rθ = 20m
2r + rθ = 20

Question 7.
If the angles of a triangle are in the ratio 1:2:3, then the smallest angle in radian is
(A) \(\frac{\pi}{3}\)
(B) \(\frac{\pi}{6}\)
(C) \(\frac{\pi}{2}\)
(D) \(\frac{\pi}{9}\)
Answer:
(B) \(\frac{\pi}{6}\)

Question 8.
A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find the ratio of their areas?
(A) 5:1
(B) 4:5
(C) 5:4
(D) 3:4
Answer:
(B) 4:5

Question 9.
Find the measure of the angle between hour- hand and the minute hand of a clock at twenty minutes past two.
(A) 50°
(B) 60°
(C) 54°
(D) 65°
Answer:
(A) 50°

Question 10.
The central angle of a sector of circle of area 9π sq.cm is 60°, the perimeter of the sector is
(A) π
(B) 3 + π
(C) 6 + π
(D) 6
Answer:
(C) 6 + π

II. Answer the following.

Question 1.
Find the number of sides of a regular polygon, if each of its interior angles is \(\frac{3 \pi^{c}}{4}\).
Solution:
Each interior angle of a regular polygon
= \(\frac{3 \pi}{4}=\left(\frac{3 \pi}{4} \times \frac{180}{\pi}\right)^{\circ}\) = 135°
Interior angle + Exterior angle = 180°
∴ Exterior angle = 180° – 135° = 45°
Let the number of sides of the regular polygon be n.
But in a regular polygon, exterior angle = \(\frac{360^{\circ}}{\text { no.of sides }}\)
∴ 45° = \(\frac{360^{\circ}}{\mathrm{n}}\)
∴ n = \(\frac{360^{\circ}}{45^{\circ}}\) = 8
∴ Number of sides of a regular polygon = 8.

Question 2.
Two circles each of radius 7 cm, intersect each other. The distance between their centres is 7√2 cm. Find the area common to both the circles.
Solution:
Let O and O₁ be the centres of two circles intersecting each other at A and B.
Then OA = OB = O₁A = O₁B = 7 cm
and OO₁ = 7√2 cm
OO₁² = 98 ………………(i)
Since OA² + O₁A² = 7²
= 98
= OO₁² …..[ from (i)]
m∠OAO₁ = 90°
□ OAO₁B is a square.
m∠AOB = m∠AO₁B = 90°

A(□ OAO₁B) = (side)² = (7)² = 49 sq.cm
∴ Required area = area of shaded portion = A(sector OAB) + A(sector O₁AB)) – A(□ OAO₁B)

Question 3.
∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.
Solution:
Let ‘O’ be the centre of the circle drawn on QR as a diameter.
Let the circle intersect seg PQ and seg PR at points M and N respectively.
Since l(OQ) = l(OM),
m∠OM Q = m∠OQM = 60°
m∠MOQ = 60°
Similarly, m∠NOR = 60°
Given, QR =18 cm.
r = 9 cm

θ = 60° = (60 x \(\frac{\pi}{180}\))^c
= \(\left(\frac{\pi}{3}\right)^{c}\)
∴ l(arc MN) = S = rθ = 9 x \(\frac{\pi}{3}\) = 3π cm.

Question 4.
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm.
Solution:
Let S be the length of the arc and r be the radius of the circle.
θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
S = 37.4 cm
Since S = rθ,

Question 5.
A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?
Solution:
S = 10 cm and r = 4 cm
Since S = rθ,
10 = 4 x θ

Question 6.
If two arcs of the same length in two circles subtend angles 65° and 110° at the centre. Find the ratio of their radii.
Solution:
Let r₁ and r₂ be the radii of the two circles and let their arcs of same length S subtend angles of 65° and 110° at their centres.
Angle subtended at the centre of the first circle,

Angle subtended at the centre of the second circle,

Question 7.
The area of a circle is 81TH sq.cm. Find the length of the arc subtending an angle of 300° at the centre and also the area of corresponding sector.
Solution:
Area of circle = πr²
But area is given to be 81 n sq.cm
∴ πr² = 81π
∴ r2 = 81
∴ r = 9 cm
θ = 300° = \(=\left(300 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{5 \pi}{3}\right)^{\mathrm{c}}\)
Since S = rθ
S = 9 x \(\frac{5 \pi}{3}\) = 15π cm
Area of sector = \(\frac{1}{2}\) x r x S
= \(\frac{1}{2}\) x 9 x 15π = \(\frac{135 \pi}{2}\) sq.cm

Question 8.
Show that minute-hand of a clock gains 5° 30′ on the hour-hand in one minute.
Solution:
Angle made by hour-hand in one minute
\(=\frac{360^{\circ}}{12 \times 60}=\left(\frac{1}{2}\right)^{\circ}\)
Angle made by minute-hand in one minute = \(\frac{360^{\circ}}{60}\) = 6°
∴ Gain by minute-hand on the hour-hand in one minute
= \(6^{\circ}-\left(\frac{1}{2}\right)^{\circ}=\left(5 \frac{1}{2}\right)^{\circ}\) = 5°30′
[Note: The question has been modified.]

Question 9.
A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.
Solution:
r = 1km = 1000m
l(Arc covered by train in 30 seconds)
= 30 x \(\frac{36000}{60 \times 60}\)m
∴ S = 300 m
Since S = rθ,
300 = 1000 x θ

= (17.18)°
= 17° +(0.18)°
= 17° + (0.18 x 60)’ = 17° + (10.8)’
∴ θ = 17°11′(approx.)

Question 10.
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:

Let ‘O’ be the centre of the circle and AB be the chord of the circle.
Here, d = 40 cm
∴ r = \(\frac{40}{2}\) = 20 cm
Since OA = OB = AB,
∆OAB is an equilateral triangle.
The angle subtended at the centre by the minor
arc AOB is θ = 60° = \(\left(60 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{3}\right)^{c}\)
= l(minor arc of chord AB) = rθ = 20 x \(\frac{\pi}{3}\)
= \(\frac{20 \pi}{3}\) cm

Question 11.
The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radians.
Solution:
Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0
∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360°
… [Sum of the angles of a quadrilateral is 360°]
∴ 4a = 360°
∴ a = 90°
According to the given condition, the greatest angle is double the least,
∴ a + 3d = 2.(a – 3d)
∴ 90° + 3d = 2.(90° – 3d)
∴ 90° + 3d = 180° – 6d 9d = 90°
∴ d = 10°
∴ The measures of the angles in degrees are
a – 3d = 90° – 3(10°) = 90° – 30° = 60°,
a – d = 90° – 10° = 80°,
a + d = 90°+ 10°= 100°,
a + 3d = 90° + 3(10°) = 90° + 30° = 120°

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.3

Question 1.
Find the joint equation of the pair of lines:
(i) Through the point (2, -1) and parallel to lines represented by 2x² + 3xy – 9y² = 0
Solution:
The combined equation of the given lines is
2x² + 3 xy – 9y² = 0
i.e. 2x² + 6xy – 3xy – 9y² = 0
i.e. 2x(x + 3y) – 3y(x + 3y) = 0
i.e. (x + 3y)(2x – 3y) = 0
∴ their separate equations are
x + 3y = 0 and 2x – 3y = 0
∴ their slopes are m₁ = \(\frac{-1}{3}\) and m₂ = \(\frac{-2}{-3}=\frac{2}{3}\).
The slopes of the lines parallel to these lines are m₁ and m₂, i.e. \(-\frac{1}{3}\) and \(\frac{2}{3}\).
∴ the equations of the lines with these slopes and through the point (2, -1) are
y + 1 = \(-\frac{1}{3}\) (x – 2) and y + 1 = \(\frac{2}{3}\)(x – 2)
i.e. 3y + 3= -x + 2 and 3y + 3 = 2x – 4
i.e. x + 3y + 1 = 0 and 2x – 3y – 7 = 0
∴ the joint equation of these lines is
(x + 3y + 1)(2x – 3y – 7) = 0
∴ 2x² – 3xy – 7x + 6xy – 9y² – 21y + 2x – 3y – 7 = 0
∴ 2x² + 3xy – 9y² – 5x – 24y – 7 = 0.

(ii) Through the point (2, -3) and parallel to lines represented by x² + xy – y² = 0
Solution:
Comparing the equation
x² + xy – y² = 0 … (1)
with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 1, b = -1
Let m₁ and m₂ be the slopes of the lines represented by (1).

The slopes of the lines parallel to these lines are m₁ and m₂.
∴ the equations of the lines with these slopes and through the point (2, -3) are
y + 3 = m₁(x – 2) and y + 3 = m₂(x – 2)
i.e. m₁(x – 2) – (y + 3) = 0 and m₂(x – 2) – (y + 3) = 0
∴ the joint equation of these lines is
[m₁(x – 2) – (y + 3)][m₂(x – 2) – (y + 3)] = 0
∴ m₁m₂(x – 2)² – m₁(x – 2)(y + 3) – m₂(x – 2)(y + 3) + (y + 3)² = o
∴ m₁m₂(x – 2)² – (m₁ + m₂)(x – 2)(y + 3) + (y + 3)³ = 0
∴ -(x – 2)² – (x – 2)(y + 3) + (y + 3)² = 0 …… [By (2)]
∴ (x – 2)² + (x – 2)(y + 3) – (y + 3)² = 0
∴ (x² – 4x + 4) + (xy + 3x – 2y – 6) – (y² + 6y + 9) = 0
∴ x² – 4x + 4 + xy + 3x – 2y – 6 – y² – 6y – 9 = 0
∴ x² + xy – y² – x – 8y – 11 = 0.

Question 2.
Show that equation x² + 2xy+ 2y² + 2x + 2y + 1 = 0 does not represent a pair of lines.
Solution:
Comparing the equation
x² + 2xy + 2y² + 2x + 2y + 1 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 1, h = 1, b = 2, g = 1, f = 1, c = 1.
The given equation represents a pair of lines, if
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|\) = 0 and h² – ab ≥ 0
Now, D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lll}
1 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 1 (2 – 1) – 1(1 – 1) + 1 (1 – 2)
= 1 – 0 – 1 = 0
and h² – ab = (1)² – 1(2) = -1 < 0
∴ given equation does not represent a pair of lines.

Question 3.
Show that equation 2x² – xy – 3y² – 6x + 19y – 20 = 0 represents a pair of lines.
Solution:
Comparing the equation
2x² – xy – 3y² – 6x + 19y – 20 = 0
with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = \(-\frac{1}{2}\), b = -3, g = -3, f = \(\frac{19}{2}\), c = -20.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
2 & -\frac{1}{2} & -3 \\
-\frac{1}{2} & -3 & \frac{19}{2} \\
-3 & \frac{19}{2} & -20
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & -1 & -6 \\
-1 & -6 & 19 \\
-6 & 19 & -40
\end{array}\right|\)
= \(\frac{1}{8}\)[4(240 – 361) + 1(40 + 114) – 6(-19 – 36)]
= \(\frac{1}{8}\)[4(-121) + 154 – 6(-55)]
= \(\frac{11}{8}\)[4(-11) + 14 – 6(-5)]
= \(\frac{1}{8}\)(-44 + 14 + 30) = 0
Also h² – ab = \(\left(-\frac{1}{2}\right)^{2}\) – 2(-3) = \(\frac{1}{4}\) + 6 = \(\frac{25}{4}\) > 0
∴ the given equation represents a pair of lines.

Question 4.
Show the equation 2x² + xy – y² + x + 4y – 3 = 0 represents a pair of lines. Also find the acute angle between them.
Solution:
Comparing the equation
2x² + xy — y² + x + 4y — 3 = 0 with
ax² + 2hxy + by² + 2gx + 2fy + c — 0, we get,
a = 2, h = \(\frac{1}{2}\), b = -1, g = \(\frac{1}{2}\), f = 2, c = – 3.
∴ D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{lrr}
2 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{2} & -1 & 2 \\
\frac{1}{2} & 2 & -3
\end{array}\right|\)
Taking \(\frac{1}{2}\) common from each row, we get,
D = \(\frac{1}{8}\left|\begin{array}{rrr}
4 & 1 & 1 \\
1 & -2 & 4 \\
1 & 4 & -6
\end{array}\right|\)
= \(\frac{1}{8}\)[4(12 —16) — 1( —6 — 4) + 1(4 + 2)]
= \(\frac{1}{8}\)[4( – 4) – 1(-10) + 1(6)]
= \(\frac{1}{8}\)(—16 + 10 + 6) = 0
Also, h² – ab = \(\left(\frac{1}{2}\right)^{2}\) – 2(-1) = \(\frac{1}{4}\) + 2 = \(\frac{9}{4}\) > 0
∴ the given equation represents a pair of lines. Let θ be the acute angle between the lines
∴ tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\)

Question 5.
Find the separate equation of the lines represented by the following equations :
(i) (x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0
Solution:
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)2 = 0
∴ (x – 2)² – 2(x – 2)(y + 1) – (x – 2)(y + 1) + 2(y + 1)² = 0
∴ (x – 2) [(x – 2) – 2(y + 1)] – (y + 1)[(x – 2) – 2(y + 1)] = 0
∴ (x – 2)(x – 2 – 2y – 2) – (y + 1)(x – 2 – 2y – 2) = 0
∴ (x – 2)(x – 2y – 4) – (y + 1)(x – 2y – 4) = 0
∴ (x – 2y – 4)(x – 2 – y – 1) = 0
∴ (x – 2y – 4)(x – y – 3) = 0
∴ the separate equations of the lines are
x – 2y – 4 = 0 and x – y – 3 = 0.
Alternative Method :
(x – 2)² – 3(x – 2)(y + 1) + 2(y + 1)² = 0 … (1)
Put x – 2 = X and y + 1 = Y
∴ (1) becomes,
X² – 3XY + 2Y² = 0
∴ X² – 2XY – XY + 2Y² = 0
∴ X(X – 2Y) – Y(X – 2Y) = 0
∴ (X – 2Y)(X – Y) = 0
∴ the separate equations of the lines are
∴ X – 2Y = 0 and X – Y = 0
∴ (x – 2) – 2(y + 1) = 0 and (x – 2) – (y +1) = 0
∴ x – 2y – 4 = 0 and x – y – 3 = 0.

(ii) 10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0
Solution:
10(x + 1)² + (x + 1)( y – 2) – 3(y – 2)² = 0 …(1)
Put x + 1 = X and y – 2 = Y
∴ (1) becomes
10x² + xy – 3y² = 0
10x² + 6xy – 5xy – 3y² = 0
2x(5x + 3y) – y(5x + 3y) = 0
(2x – y)(5x + 3y) = 0
5x + 3y = 0 and 2x – y = 0
5x + 3y = 0
5(x + 1) + 3(y – 2) = 0
5x + 5 + 3y – 6 = 0
∴ 5x + 3y – 1 = 0
2x – y = 0
2(x + 1) – (y – 2) = 0
2x + 2 – y + 2 = 0
∴ 2x – y + 4 = 0

Question 6.
Find the value of k if the following equations represent a pair of lines :
(i) 3x² + 10xy + 3y² + 16y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 3, h = 5, b = 3, g = 0, f= 8, c = k.
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (3)(3)(k) + 2(8)(0)(5) – 3(8)² – 3(0)² – k(5)² = 0
∴ 9k + 0 – 192 – 0 – 25k = 0
∴ -16k – 192 = 0
∴ – 16k = 192
∴ k= -12.

(ii) kxy + 10x + 6y + 4 = 0
Solution:
Comparing the given equation with
ax² + 2 hxy + by² + 2gx + 2fy + c = 0,
we get, a = 0, h = \(\frac{k}{2}\), b = 0, g = 5, f = 3, c = 4
Now, given equation represents a pair of lines.
∴ abc + 2fgh – af² – bg² – ch² = 0
∴ (0)(0)(4) + 2(3)(5)\(\left(\frac{k}{2}\right)\) – 0(3)² – 0(5)² – 4\(\left(\frac{k}{2}\right)^{2}\) = 0
∴ 0 + 15k – 0 – 0 – k² = 0
∴ 15k – k² = 0
∴ -k(k – 15) = 0
∴ k = 0 or k = 15.
If k = 0, then the given equation becomes
10x + 6y + 4 = 0 which does not represent a pair of lines.
∴ k ≠ o
Hence, k = 15.

(iii) x² + 3xy + 2y² + x – y + k = 0
Solution:
Comparing the given equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
we get, a = 1, h = \(\frac{3}{2}\), b = 2, g = \(\frac{1}{2}\), f= \(-\frac{1}{2}\), c = k.
Now, given equation represents a pair of lines.


∴ 2(8k – 1) – 3(6k + 1) + 1(-3 – 4) = 0
∴ 16k – 2 – 18k – 3 – 7 = 0
∴ -2k – 12 = 0
∴ -2k = 12 ∴ k = -6.

Question 7.
Find p and q if the equation px² – 8xy + 3y² + 14x + 2y + q = 0 represents a pair of perpendicular lines.
Solution:
The given equation represents a pair of lines perpendicular to each other
∴ (coefficient of x²) + (coefficient of y²) = 0
∴ p + 3 = 0 p = -3
With this value of p, the given equation is
– 3x² – 8xy + 3y² + 14x + 2y + q = 0.
Comparing this equation with
ax² + 2hxy + by² + 2gx + 2fy + c = 0, we have,
a = -3, h = -4, b = 3, g = 7, f = 1 and c = q.
D = \(\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=\left|\begin{array}{rrr}
-3 & -4 & 7 \\
-4 & 3 & 1 \\
7 & 1 & q
\end{array}\right|\)
= -3(3q – 1) + 4(-4q – 7) + 7(-4 – 21)
= -9q + 3 – 16q – 28 – 175
= -25q – 200= -25(q + 8)
Since the given equation represents a pair of lines, D = 0
∴ -25(q + 8) = 0 ∴ q= -8.
Hence, p = -3 and q = -8.

Question 8.
Find p and q if the equation 2x² + 8xy + py² + qx + 2y – 15 = 0 represents a pair of parallel lines.
Solution:
The given equation is
2x² + 8xy + py² + qx + 2y – 15 = 0
Comparing it with ax² + 2hxy + by² + 2gx + 2fy + c = 0, we get,
a = 2, h = 4, b = p, g = \(\frac{q}{2}\), f = 1, c = – 15
Since the lines are parallel, h² = ab
∴ (4)² = 2p ∴ P = 8
Since the given equation represents a pair of lines

i.e. – 242 + 240 + 2q + 2q – 2q² = 0
i.e. -2q² + 4q – 2 = 0
i.e. q² – 2q + 1 = 0
i.e. (q – 1)² = 0 ∴ q – 1 = 0 ∴ q = 1.
Hence, p = 8 and q = 1.

Question 9.
Equations of pairs of opposite sides of a parallelogram are x² – 7x+ 6 = 0 and y² – 14y + 40 = 0. Find the joint equation of its diagonals.
Solution:
Let ABCD be the parallelogram such that the combined equation of sides AB and CD is x² – 7x + 6 = 0 and the combined equation of sides BC and AD is y² – 14y + 40 = 0.
The separate equations of the lines represented by x² – 7x + 6 = 0, i.e. (x – 1)(x – 6) = 0 are x – 1 = 0 and x – 6 = 0.
Let equation of the side AB be x – 1 = 0 and equation of side CD be x – 6 = 0.
The separate equations of the lines represented by y² – 14y + 40 = 0, i.e. (y – 4)(y – 10) = 0 are y – 4 = 0 and y – 10 = 0.
Let equation of the side BC be y – 4 = 0 and equation of side AD be y – 10 = 0.

Coordinates of the vertices of the parallelogram are A(1, 10), B(1, 4), C(6, 4) and D(6, 10).
∴ equation of the diagonal AC is
\(\frac{y-10}{x-1}\) = \(\frac{10-4}{1-6}\) = \(\frac{6}{-5}\)
∴ -5y + 50 = 6x – 6
∴ 6x + 5y – 56 = 0
and equation of the diagonal BD is
\(\frac{y-4}{x-1}\) = \(\frac{4-10}{1-6}\) = \(\frac{-6}{-5}\) = \(\frac{6}{5}\)
∴ 5y – 20 = 6x – 6
∴ 6x – 5y + 14 = 0
Hence, the equations of the diagonals are 6x + 5y – 56 = 0 and 6x – 5y + 14 = 0.
∴ the joint equation of the diagonals is (6x + 5y – 56)(6x – 5y + 14) = 0
∴ 36x² – 30xy + 84x + 30xy – 25y² + 70y – 336x + 280y – 784 = 0
∴ 36x² – 25y² – 252x + 350y – 784 = 0.

Question 10.
∆OAB is formed by lines x² – 4xy + y² = 0 and the line 2x + 3y – 1 = 0. Find the equation of the median of the triangle drawn from O.
Solution:

Let D be the midpoint of seg AB where A is (x₁, y₁) and B is (x₂, y₂).
Then D has coordinates \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\).
The joint (combined) equation of the lines OA and OB is x² – 4xy + y² = 0 and the equation of the line AB is 2x + 3y – 1 = 0.
∴ points A and B satisfy the equations 2x + 3y – 1 = 0
and x² – 4xy + y² = 0 simultaneously.
We eliminate x from the above equations, i.e.,
put x = \(\frac{1-3 y}{2}\) in the equation x² – 4xy + y² = 0, we get,
∴ \(\left(\frac{1-3 y}{2}\right)^{2}\) – 4\(\left(\frac{1-3 y}{2}\right)\)y + y² = 0
∴ (1 – 3y)² – 8(1 – 3y)y + 4y² = 0
∴1 – 6y + 9y² – 8y + 24y² + 4y² = 0
∴ 37y² – 14y + 1 = 0
The roots y₁ and y₂ of the above quadratic equation are the y-coordinates of the points A and B.
∴ y₁ + y₂ = \(\frac{-b}{a}=\frac{14}{37}\)
∴ y-coordinate of D = \(\frac{y_{1}+y_{2}}{2}=\frac{7}{37}\).
Since D lies on the line AB, we can find the x-coordinate of D as
2x + 3\(\left(\frac{7}{37}\right)\) – 1 = 0
∴ 2x = 1 – \(\frac{21}{37}=\frac{16}{37}\)
∴ x = \(\frac{8}{37}\)
∴ D is (8/37, 7/37)
∴ equation of the median OD is \(\frac{x}{8 / 37}=\frac{y}{7 / 37}\),
i.e., 7x – 8y = 0.

Question 11.
Find the co-ordinates of the points of intersection of the lines represented by x² – y² – 2x + 1 = 0.
Solution:
Consider, x² – y² – 2x + 1 = 0
∴ (x² – 2x + 1) – y² = 0
∴ (x – 1)² – y² = 0
∴ (x – 1 + y)(x – 1 – y) = 0
∴ (x + y – 1)(x – y – 1) = 0
∴ separate equations of the lines are
x + y – 1 = 0 and x – y +1 = 0.
To find the point of intersection of the lines, we have to solve
x + y – 1 = 0 … (1)
and x – y + 1 = 0 … (2)
Adding (1) and (2), we get,
2x = 0 ∴ x = 0
Substituting x = 0 in (1), we get,
0 + y – 1 = 0 ∴ y = 1
∴ coordinates of the point of intersection of the lines are (0, 1).

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.2

Question 1.
Show that lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.
Solution:
Comparing the equation 3x² – 4 xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = -4, b = -3 Since a + b = 3 + (-3) = 0, the lines represented by 3x² – 4xy – 3y² = 0 are perpendicular to each other.

Question 2.
Show that lines represented by x² + 6xy + gy²= 0 are coincident.
Question is modified.
Show that lines represented by x² + 6xy + 9y²= 0 are coincident.
Solution:
Comparing the equation x² + 6xy + 9y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 1, 2h = 6, i.e. h = 3 and b = 9
Since h² – ab = (3)² – 1(9)
= 9 – 9 = 0, .
the lines represented by x² + 6xy + 9y² = 0 are coincident.

Question 3.
Find the value of k if lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other.
Solution:
Comparing the equation kx² + 4xy – 4y² = 0 with ax² + 2hxy + by² = 0, we get,
a = k, 2h = 4, b = -4
Since lines represented by kx² + 4xy – 4y² = 0 are perpendicular to each other,
a + b = 0
∴ k – 4 = 0 ∴ k = 4.

Question 4.
Find the measure of the acute angle between the lines represented by:
(i) 3x² – 4\(\sqrt {3}\)xy + 3y² = 0
Solution:
Comparing the equation 3x² – 4\(\sqrt {3}\)xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 3, 2h = -4\(\sqrt {3}\), i.e. h = -24\(\sqrt {3}\) and b = 3
Let θ be the acute angle between the lines.

∴ θ = 30°.

(ii) 4x² + 5xy + y² = 0
Solution:
Comparing the equation 4x² + 5xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 4, 2h = 5, i.e. h = \(\frac{5}{2}\) and b = 1.
Let θ be the acute angle between the lines.

(iii) 2x² + 7xy + 3y² = 0
Solution:
Comparing the equation
2x² + 7xy + 3y² = 0 with
ax² + 2hxy + by² = 0, we get,
a = 2, 2h = 7 i.e. h = \(\frac{7}{2}\) and b = 3
Let θ be the acute angle between the lines.


tanθ = 1
∴ θ = tan 1 = 45°
∴ θ = 45°

(iv) (a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0
Solution:
Comparing the equation
(a² – 3b²)x² + 8abxy + (b² – 3a²)y² = 0, with
Ax² + 2Hxy + By² = 0, we have,
A = a² – 3b², H = 4ab, B = b² – 3a².
∴ H² – AB = 16a²b² – (a² – 3b²)(b² – 3a²)
= 16a²b² + (a² – 3b²)(3a² – b²)
= 16a²b² + 3a⁴ – 10a²b² + 3b⁴
= 3a⁴ + 6a²b² + 3b⁴
= 3(a⁴ + 2a²b² + b⁴)
= 3 (a² + b²)²
∴ \(\sqrt{H^{2}-A B}\) = \(\sqrt {3}\) (a² + b²)
Also, A + B = (a² – 3b²) + (b² – 3a²)
= -2 (a² + b²)
If θ is the acute angle between the lines, then
tan θ = \(\left|\frac{2 \sqrt{H^{2}-A B}}{A+B}\right|=\left|\frac{2 \sqrt{3}\left(a^{2}+b^{2}\right)}{-2\left(a^{2}+b^{2}\right)}\right|\)
= \(\sqrt {3}\) = tan 60°
∴ θ = 60°

Question 5.
Find the combined equation of lines passing through the origin each of which making an angle of 30° with the line 3x + 2y – 11 = 0
Solution:
The slope of the line 3x + 2y – 11 = 0 is m₁ = \(-\frac{3}{2}\) .
Let m be the slope of one of the lines making an angle of 30° with the line 3x + 2y – 11 = 0.
The angle between the lines having slopes m and m1 is 30°.

On squaring both sides, we get,
\(\frac{1}{3}=\frac{(2 m+3)^{2}}{(2-3 m)^{2}}\)
∴ (2 – 3m)² = 3 (2m + 3)²
∴ 4 – 12m + 9m² = 3(4m² + 12m + 9)
∴ 4 – 12m + 9m² = 12m² + 36m + 27
3m² + 48m + 23 = 0
This is the auxiliary equation of the two lines and their joint equation is obtained by putting m = \(\frac{y}{x}\).
∴ the combined equation of the two lines is
3\(\left(\frac{y}{x}\right)^{2}\) + 48\(\left(\frac{y}{x}\right)\) + 23 = 0
∴ \(\frac{3 y^{2}}{x^{2}}+\frac{48 y}{x}\) + 23 = 0
∴ 3y² + 48xy + 23x² = 0
∴ 23x² + 48xy + 3y² = 0.

Question 6.
If the angle between lines represented by ax² + 2hxy + by² = 0 is equal to the angle between lines represented by 2x² – 5xy + 3y² = 0 then show that 100(h² – ab) = (a + b)².
Solution:
The acute angle θ between the lines ax² + 2hxy + by² = 0 is given by
tan θ = \(\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|\) ..(1)
Comparing the equation 2x² – 5xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 2, 2h= -5, i.e. h = \(-\frac{5}{2}\) and b = 3
Let ∝ be the acute angle between the lines 2x² – 5xy + 3y² = 0.

This is the required condition.

Question 7.
Find the combined equation of lines passing through the origin and each of which making angle 60° with the Y- axis.
Solution:

Let OA and OB be the lines through the origin making an angle of 60° with the Y-axis.
Then OA and OB make an angle of 30° and 150° with the positive direction of X-axis.
∴ slope of OA = tan 30° = \(\frac{1}{\sqrt{3}}\)
∴ equation of the line OA is
y = \(\frac{1}{\sqrt{3}}\) = x, i.e. x – \(\sqrt {3}\)y = 0
Slope of OB = tan 150° = tan (180° – 30°)
= tan 30° = \(-\frac{1}{\sqrt{3}}\)
∴ equation of the line OB is
y = \(-\frac{1}{\sqrt{3}}\)x, i.e. x + \(\sqrt {3}\) y = 0
∴ required combined equation is
(x – \(\sqrt {3}\)y)(x + \(\sqrt {3}\)y) = 0
i.e. x² – 3y² = 0.

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 4 Pair of Straight Lines Ex 4.1 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 4 Pair of Straight Lines Ex 4.1

Question 1.
Find the combined equation of the following pairs of lines:
(i) 2x + y = 0 and 3x – y = 0
Solution:
The combined equation of the lines 2x + y = 0 and 3x – y = 0 is
(2x + y)( 3x – y) = 0
∴ 6x² – 2xy + 3xy – y² = 0
∴ 6x² – xy – y² = 0.

(ii) x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
The combined equation of the lines x + 2y – 1 = 0 and x – 3y + 2 = 0 is
(x + 2y – 1)(x – 3y + 2) = 0
∴ x² – 3xy + 2x + 2xy – 6y² + 4y – x + 3y – 2 = 0
∴ x² – xy – 6y² + x + 7y – 2 = 0.

(iii) Passing through (2, 3) and parallel to the co-ordinate axes.
Solution:
Equations of the coordinate axes are x = 0 and y = 0.
∴ the equations of the lines passing through (2, 3) and parallel to the coordinate axes are x = 2 and
i.e. x – 2 = 0 and y – 3 = 0.
∴ their combined equation is
(x – 2)(y – 3) = 0.
∴ xy – 3x – 2y + 6 = 0.

(iv) Passing through (2, 3) and perpendicular to lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0
Solution:
Let L₁ and L₂ be the lines passing through the point (2, 3) and perpendicular to the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 respectively.
Slopes of the lines 3x + 2y – 1 = 0 and x – 3y + 2 = 0 are \(\frac{-3}{2}\) and \(\frac{-1}{-3}=\frac{1}{3}\) respectively.
∴ slopes of the lines L₁ and L₂ are \(\frac{2}{3}\) and -3 respectively.
Since the lines L₁ and L₂ pass through the point (2, 3), their equations are
y – 3 = \(\frac{2}{3}\)(x – 2) and y – 3 = -3 (x – 2)
∴ 3y – 9 = 2x – 4 and y – 3= -3x + 6
∴ 2x – 3y + 5 = 0 and 3x – y – 9 = 0
∴ their combined equation is
(2x – 3y + 5)(3x + y – 9) = 0
∴ 6x² + 2xy – 18x – 9xy – 3y² + 27y + 15x + 5y – 45 = 0
∴ 6x² – 7xy – 3y² – 3x + 32y – 45 = 0.

(v) Passsing through (-1, 2),one is parallel to x + 3y – 1 = 0 and the other is perpendicular to 2x – 3y – 1 = 0.
Solution:
Let L₁ be the line passing through (-1, 2) and parallel to the line x + 3y – 1 = 0 whose slope is –\(\frac{1}{3}\).
∴ slope of the line L₁ is –\(\frac{1}{3}\)
∴ equation of the line L₁ is
y – 2 = –\(\frac{1}{3}\)(x + 1)
∴ 3y – 6 = -x – 1
∴ x + 3y – 5 = 0
Let L₂ be the line passing through (-1, 2) and perpendicular to the line 2x – 3y – 1 = 0
whose slope is \(\frac{-2}{-3}=\frac{2}{3}\).
∴ slope of the line L₂ is –\(\frac{3}{2}\)
∴ equation of the line L₂ is
y – 2= –\(\frac{3}{2}\)(x + 1)
∴ 2y – 4 = -3x – 3
∴ 3x + 2y – 1 = 0
Hence, the equations of the required lines are
x + 3y – 5 = 0 and 3x + 2y – 1 = 0
∴ their combined equation is
(x + 3y – 5)(3x + 2y – 1) = 0
∴ 3x² + 2xy – x + 9xy + 6y² – 3y – 15x – 10y + 5 = 0
∴ 3x² + 11xy + 6y² – 16x – 13y + 5 = 0

Question 2.
Find the separate equations of the lines represented by following equations:
(i) 3y² + 7xy = 0
Solution:
3y² + 7xy = 0
∴ y(3y + 7x) = 0
∴ the separate equations of the lines are y = 0 and 7x + 3y = 0.

(ii) 5x² – 9y² = 0
Solution:
5x² – 9y² = 0
∴ (\(\sqrt {5}\) x)² – (3y)² = 0
∴ (\(\sqrt {5}\)x + 3y)(\(\sqrt {5}\)x – 3y) = 0
∴ the separate equations of the lines are
\(\sqrt {5}\)x + 3y = 0 and \(\sqrt {5}\)x – 3y = 0.

(iii) x² – 4xy = 0
Solution:
x² – 4xy = 0
∴ x(x – 4y) = 0
∴ the separate equations of the lines are x = 0 and x – 4y = 0

(iv) 3x² – 10xy – 8y² = 0
Solution:
3x² – 10xy – 8y² = 0
∴ 3x² – 12xy + 2xy – 8y² = 0
∴ 3x(x – 4y) + 2y(x – 4y) = 0
∴ (x – 4y)(3x +2y) = 0
∴ the separate equations of the lines are x – 4y = 0 and 3x + 2y = 0.

(v) 3x² – \(2 \sqrt{3}\) xy – 3y² = 0
Solution:
3x² – 2\(\sqrt {3}\)xy – 3y2 = 0
∴ 3x² – 3\(\sqrt {3}\)xy + \(\sqrt {3}\)xy – 3y² = 0
∴ 3x(x – \(\sqrt {3}\)y) + \(\sqrt {3}\)y(x – \(\sqrt {3}\)y) = 0
∴ (x – \(\sqrt {3}\)y)(3x + \(\sqrt {3}\)y) = 0
∴ the separate equations of the lines are
∴ x – \(\sqrt {3}\)y = 0 and 3x + \(\sqrt {3}\)y = 0.

(vi) x² + 2(cosec ∝)xy + y² = 0
Solution:
x² + 2 (cosec ∝)xy – y² = 0
i.e. y² + 2(cosec∝)xy + x² = 0
Dividing by x², we get,

∴ the separate equations of the lines are
(cosec ∝ – cot ∝)x + y = 0 and (cosec ∝ + cot ∝)x + y = 0.

(vii) x² + 2xy tan ∝ – y² = 0
Solution:
x² + 2xy tan ∝ – y² = 0
Dividind by y²

The separate equations of the lines are
(sec∝ – tan ∝)x + y = 0 and (sec ∝ + tan ∝)x – y = 0

Question 3.
Find the combined equation of a pair of lines passing through the origin and perpendicular
to the lines represented by following equations :
(i) 5x² – 8xy + 3y² = 0
Solution:
Comparing the equation 5x² – 8xy + 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = -8, b = 3
Let m₁ and m₂ be the slopes of the lines represented by 5x² – 8xy + 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{8}{3}\)
amd m₁m₂ = \(\frac{a}{b}=\frac{5}{3}\) …(1)
Now required lines are perpendicular to these lines
∴ their slopes are -1 /m₁ and -1/m₂ Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{8}{3}\)xy + \(\frac{5}{3}\)y² = 0 … [By (1)]
∴ x² + 8xy + 5y\(\frac{8}{3}\) = 0

(ii) 5x² + 2xy – 3y² = 0
Solution:
Comparing the equation 5x² + 2xy – 3y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 5, 2h = 2, b = -3
Let m₁ and m₂ be the slopes of the lines represented by 5x² + 2xy – 3y² = 0
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-2}{-3}=\frac{2}{3}\) and m₁m₂ = \(\frac{a}{b}=\frac{5}{-3}\) ..(1)
Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\)
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{\mathrm{~m}_{1}}\)x and y = \(\frac{-1}{\mathrm{~m}_{2}}\)x
i.e. m₁y = -x amd m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
∴ (x + m₁y)(x + m₂y) = 0
x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² + \(\frac{2}{3}\)xy – \(\frac{5}{3}\)y = 0 …[By (1)]
∴ 3x² + 2xy – 5y² = 0

(iii) xy + y² = 0
Solution:
Comparing the equation xy + y² = 0 with ax² + 2hxy + by² = 0, we get,
a = 0, 2h = 1, b = 1
Let m₁ and m₂ be the slopes of the lines represented by xy + y² = 0

Now required lines are perpendicular to these lines
∴ their slopes are \(\frac{-1}{m_{1}}\) and \(\frac{-1}{m_{2}}\).
Since these lines are passing through the origin, their separate equations are
y = \(\frac{-1}{m_{1}}\)x and y = \(\frac{-1}{m_{2}}\)x
i.e. m₁y = -x and m₂y = -x
i.e. x + m₁y = 0 and x + m₂y = 0
∴ their combined equation is
(x + m₁y) (x + m₂y) = 0
∴ x² + (m₁ + m₂)xy + m₁m₂y² = 0
∴ x² – xy = 0.y² = 0 … [By (1)]
∴ x² – xy = 0.
Alternative Method :
Consider xy + y² = 0
∴ y(x + y) = 0
∴ separate equations of the lines are y = 0 and
3x² + 8xy + 5y² = 0.
x + y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ = 0 and m₂ = -1
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\)
Since, m₁ = 0, \(-\frac{1}{m_{1}}\) does not exist.
Also, m₂ = -1, \(-\frac{1}{m_{2}}\) = 1
Since these lines are passing through the origin, their separate equations are x = 0 and y = x,
i.e. x – y = 0
∴ their combined equation is
x(x – y) = 0
x² – xy = 0.

(iv) 3x² – 4xy = 0
Solution:
Consider 3x² – 4xy = 0
∴ x(3x – 4y) = 0
∴ separate equations of the lines are x = 0 and 3x – 4y = 0.
Let m₁ and m₂ be the slopes of these lines.
Then m₁ does not exist and and m₁ = \(\frac{3}{4}\).
Now, required lines are perpendicular to these lines.
∴ their slopes are \(-\frac{1}{m_{1}}\) and \(-\frac{1}{m_{2}}\).
Since m₁ does not exist, \(-\frac{1}{m_{1}}\) = 0
Also m₂ = \(\frac{3}{4^{\prime}}-\frac{1}{m_{2}}=-\frac{4}{3}\)
Since these lines are passing through the origin, their separate equations are y = 0 and y = \(-\frac{4}{3}\)x,
i.e.   4x + 3y = 0
∴ their combined equation is
y(4x + 3y) = 0
∴ 4xy + 3y² = 0.

Question 4.
Find k if,
(i) the sum of the slopes of the lines represented by x² + kxy – 3y² = 0 is twice their product.
Solution:
Comparing the equation x² + kxy – 3y² = 0 with ax² + 2hxy + by² = 0, we get, a = 1, 2h = k, b = -3.
Let m₁ and m₂ be the slopes of the lines represented by x² + kxy – 3y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{(-3)}=\frac{k}{3}\)
and m₁m₂ = \(\frac{a}{b}=\frac{1}{(-3)}=-\frac{1}{3}\)
Now, m₁ + m₂ = 2(m₁m₂) ..(Given)
∴ \(\frac{k}{3}=2\left(-\frac{1}{3}\right)\) ∴ k = -2

(ii) slopes of lines represent by 3x² + kxy – y² = 0 differ by 4.
Solution:
(ii) Comparing the equation 3x² + kxy – y² = 0 with ax² + 2hxy + by² = 0, we get, a = 3, 2h = k, b = -1.
Let m₁ and m₂ be the slopes of the lines represented by 3x² + kxy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=-\frac{k}{-1}\) = k
and m₁₂ = \(\frac{a}{b}=\frac{3}{-1}\) = -3
∴ (m₁ – m₂)² = (m₁ + m₂)² – 4m₁m₂
= k² – 4 (-3)
= k² + 12 … (1)
But |m₁ – m₂| =4
∴ (m₁ – m₂)² = 16 … (2)
∴ from (1) and (2), k² + 12 = 16
∴ k² = 4 ∴ k= ±2.

(iii) slope of one of the lines given by kx² + 4xy – y² = 0 exceeds the slope of the other by 8.
Solution:
Comparing the equation kx² + 4xy – y² = 0 with ² + 2hxy + by² = 0, we get, a = k, 2h = 4, b = -1. Let m₁ and m₂ be the slopes of the lines represented by kx² + 4xy – y² = 0.
∴ m₁ + m₂ = \(\frac{-2 h}{b}=\frac{-4}{-1}\) = 4
and m₁m₂ = \(\frac{a}{b}=\frac{k}{-1}\) = -k
We are given that m₂ = m₁ + 8
m₁ + m₁ + 8 = 4
∴ 2m₁ = -4 ∴ m₁ = -2 … (1)
Also, m₁(m₁ + 8) = -k
(-2)(-2 + 8) = -k … [By(1)]
∴ (-2)(6) = -k
∴ -12= -k ∴ k = 12.

Question 5.
Find the condition that :
(i) the line 4x + 5y = 0 coincides with one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Given that 4x + 5y = 0 is one of the lines represented by ax² + 2hxy + by² = 0.
The slope of the line 4x + 5y = 0 is \(-\frac{4}{5}\).
∴ m = \(-\frac{4}{5}\) is a root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(-\frac{4}{5}\right)^{2}\) + 2h\(\left(-\frac{4}{5}\right)\) + a = 0
∴ \(\frac{16 b}{25}-\frac{8 h}{5}\) + a = 0
∴ 16b – 40h + 25a = 0
∴ 25a + 16b = 40k.
This is the required condition.

(ii) the line 3x + y = 0 may be perpendicular to one of the lines given by ax² + 2hxy + by² = 0.
Solution:
The auxiliary equation of the lines represented by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one line is perpendicular to the line 3x + y = 0
whose slope is \(-\frac{3}{1}\) = -3
∴ slope of that line = m = \(\frac{1}{3}\)
∴ m = \(\frac{1}{3}\)is the root of the auxiliary equation bm² + 2hm + a = 0.
∴ b\(\left(\frac{1}{3}\right)^{2}\) + 2h\(\left(\frac{1}{3}\right)\) + a = 0
∴ \(\frac{b}{9}+\frac{2 h}{3}\) + a = 0
∴ b + 6h + 9a = 0
∴ 9a + b + 6h = 0
This is the required condition.

Question 6.
If one of the lines given by ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0 then show that ap² + 2hpq + bq² = 0.
Solution:
To prove ap² + 2hpq + bq² = 0.
Let the slope of the pair of straight lines ax² + 2hxy + by² = 0 be m₁ and m₂
Then, m₁ + m₂ = \(\frac{-2 h}{b}\) and m₁m₂ = \(\frac{a}{b}\)
Slope of the line px + qy = 0 is \(\frac{-p}{q}\)
But one of the lines of ax² + 2hxy + by² = 0 is perpendicular to px + qy = 0

⇒ bq² + ap² = -2hpq
⇒ ap² + 2hpq + bq² = 0

Question 7.
Find the combined equation of the pair of lines passing through the origin and making an equilateral triangle with the line y = 3.
Solution:
Let OA and OB be the lines through the origin making.an angle of 60° with the line y = 3.
∴ OA and OB make an angle of 60° and 120° with the positive direction of X-axis.
∴ slope of OA = tan60° = \(\sqrt {3}\)
∴ equation of the line OA is
y = \(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x – y = 0

Slope of OB = tan 120° = tan (180° – 60°)
= -tan 60°= –\(\sqrt {3}\)
∴ equation of the line OB is
y = –\(\sqrt {3}\) x, i.e. \(\sqrt {3}\) x + y = 0
∴ required joint equation of the lines is
(\(\sqrt {3}\) x – y)(\(\sqrt {3}\) x + y) = 0
i.e. 3x² – y² = 0.

Question 8.
If slope of one of the lines given by ax² + 2hxy + by² = 0 is four times the other then show that 16h² = 25ab.
Solution:
Let m₁ and m₂ be the slopes of the lines given by ax² + 2hxy + by² = 0.
∴ m₁ + m₂ = \(-\frac{2 h}{b}\)
and m₁m₂ = \(\frac{a}{b}\)
We are given that m₂ = 4m₁

∴ 16h² = 25ab
This is the required condition.

Question 9.
If one of the lines given by ax² + 2hxy + by² = 0 bisects an angle between co-ordinate axes then show that (a + b) ² = 4h².
Solution:
The auxiliary equation of the lines given by ax² + 2hxy + by² = 0 is bm² + 2hm + a = 0.
Since one of the line bisects an angle between the coordinate axes, that line makes an angle of 45° or 135° with the positive direction of X-axis.
∴ slope of that line = tan45° or tan 135°
∴ m = tan45° = 1
or m = tan 135° = tan (180° – 45°)
= -tan 45°= -1
∴ m = ±1 are the roots of the auxiliary equation bm² + 2hm + a = 0.
∴ b(±1)² + 2h(±1) + a = 0
∴ b ± 2h + a = 0
∴ a + b = ±2h
∴ (a + b)² = 4h²
This is the required condition.