Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.1 Questions and Answers.

## Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1

Question 1.

If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.

Solution:

Let P(x, y) be any point on the required locus.

Given, A(1, 3), B(2, 1) and

PA = PB

∴ PA^{2} = PB^{2}

∴ (x – 1)^{2} + ( y – 3)^{2} = (x – 2)^{2} + (y – 1)^{2}

∴ x^{2} – 2x + 1 + y^{2} – 6y + 9 = x^{2} – 4x + 4 + y^{2} – 2y + 1

-2x – 6y + 10 = -4x – 2y + 5

∴ 2x – 4y + 5 = 0

∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.

A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B.

Solution:

Let P(x, y) be any point on the required locus.

P is equidistant from A(- 5, 2) and B(4, 1).

∴ PA = PB

∴ PA^{2} = PB^{2}

∴ (x + 5)^{2} + (y – 2)^{2} = (x – 4)^{2} + (y – 1)^{2}

∴ x^{2} + 10x + 25 + y^{2} — 4y + 4

= x^{2} – 8x + 16 + y^{2} – 2y + 1

∴ 10x – 4y + 29 = -8x – 2y + 17

∴ 18x – 2y + 12 = 0

∴ 9x – y + 6 = 0

The required equation of locus is 9x -y + 6 = 0.

Question 3.

If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.

Solution:

Let P(x, y) be any point on the required locus.

Given, A(2, 0), B(0, 3) and AP = 2BP

∴ AP^{2} = 4BP^{2}

∴ (x – 2)^{2} + (y – 0)^{2} = 4[(x – 0)^{2} + (y – 3)^{2}]

∴ x^{2} – 4x + 4 + y^{2} = 4(x^{2} + y^{2} – 6y + 9)

x^{2} – 4x + 4 + y^{2} = 4x^{2} + 4y^{2} – 24y + 36

∴ 3x^{2} + 3 y^{2} + 4x – 24y + 32 = 0

∴ The required equation of locus is

3x^{2} + 3y^{2} + 4x – 24y + 32 = 0.

[Note: Answer given in the textbook , is

‘3x^{2} + 3y^{2} + 4x + 24y + 32 = O’.

However, as per our calculation it is ‘3x^{2} + 3y^{2} + 4x – 24y + 32 = 0’.]

Question 4.

If A(4,1) and B(5,4), find the equation of the locus of point P such that PA^{2} = 3PB^{2}.

Solution:

Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA2 = 3PB2

∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)^{2} + (y – 4)^{2}]

∴ x^{2} – 8x + 16 + y^{2} – 2y + 1 = 3(x^{2} – 10x + 25 + y^{2} – 8y + 16)

∴ x^{2} – 8x + y^{2} – 2y + 17 = 3x^{2} -30x + 75 + 3y^{2} – 24y + 48

∴ 2x^{2} + 2y^{2} – 22x – 22y + 106 = 0

∴ x^{2} + y^{2} – 11x – 11y + 53 = 0

∴ The required equation of locus is

x^{2} + y^{2} – 11x – 11y + 53 = 0.

Question 5.

A(2, 4) and B(5, 8), find the equation of the

locus of point P such that PA^{2} – PB^{2} = 13.

Solution:

Let P(x, y) be any point on the required locus. Given, A(2,4), B(5, 8) and PA^{2}-PB^{2} = 13

∴ [(x -2)^{2} + (y – 4)^{2}] – [(x -5)^{2} + (y- 8)^{2}] = 13

∴ (x^{2} – 4x + 4 + y^{2} – 8y + 16) – (x^{2} – 10x + 25 + y^{2} – 16y + 64) =13

∴ x^{2} – 4x+ y^{2} – 8y + 20 – x^{2} + 10x – y^{2} + 16y – 89 = 13

∴ 6x + 8y- 69 = 13

∴ 6x + 8y – 82 = 0

∴ 3x + 4y – 41 = 0

∴ The required equation of locus is 3x + 4y- 41 = 0.

Question 6.

A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)

Solution:

Let P(x, y) be any point on the required locus. Given,

A(l, 6) and B(3, 5),

∠APB = 90°

∴ ΔAPB is a right angled triangle,

By Pythagoras theorem,

AP^{2} + PB^{2} = AB^{2} P (x,y)

∴ [(x – 1)^{2} + (y – 6)^{2}] + [(x – 3)^{2} + (y – 5)^{2}] = (1 – 3)^{2} + (6 -5)^{2}

∴ x^{2} — 2x + 1 + y^{2} — 12y + 36 + x^{2} – 6x + 9 + y^{2} – 10y + 25 = 4 + 1

∴ 2x^{2} + 2y^{2} – 8x – 22y + 66 = 0

∴ x^{2} + y^{2} – 4x – 11y + 33 = 0

∴ The required equation of locus is x^{2} + y^{2} – 4x – 11y + 33 = 0.

[Note: Answer given in the textbook is

‘3x^{2} + 4y^{2} – 4x – 11y + 33 = 0’.

However, as per our calculation it is ‘x^{2} + y^{2} – 4x – 11y + 33 = O

Question 7.

If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points

i. A(1, 3) ii. B(2,5)

Solution:

Origin is shifted to (2, 3) = (h, k)

Let the new co-ordinates be (X, Y).

x = X + handy = Y + k

x = X + 2 andy = Y + 3 …(i)

i. Given, A(x, y) = A( 1, 3)

x = X + 2 andy = Y + 3 …[From(i)]

∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0

∴ The new co-ordinates of point A are (- 1,0).

ii. Given, B(x, y) = B(2, 5)

x = X + 2 and y = Y + 3 …[From(i)]

∴ 2 = X + 2 and 5 = Y + 3

∴ X = 0 and Y = 2

∴ The new co-ordinates of point B are (0, 2).

Question 8.

If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points

i. C(5,4) ii. D(3,3)

Solution:

Origin is shifted to (1,3) = (h, k)

Let the new co-ordinates be (X, Y).

x = X + h andy = Y + k

∴ x = X+1 andy = Y + 3 …(i)

i. Given, C(X, Y) = C(5, 4)

x = X +1 andy = Y + 3 …[From(i)]

∴ x = 5 + 1 = 6 andy = 4 + 3 = 7

∴ The old co-ordinates of point C are (6, 7).

ii. Given, D(X, Y) = D(3, 3)

x = X + 1 andy = Y + 3 …[From(i)]

∴ x = 3 + 1 = 4 and y = 3 + 3 = 6

∴ The old co-ordinates of point D are (4, 6).

Question 9.

If the co-ordinates A(5, 14) change to B(8, 3) by shift of origin, find the co-ordinates of the point, where the origin is shifted.

Solution:

Let the origin be shifted to (h, k).

Given, A(x, y) = A(5,14), B(X, Y) = B(8, 3)

Since x = X + h andy = Y + k,

5 = 8 + hand 14 = 3 + k ,

∴ h = – 3 and k = 11

The co-ordinates of the point, where the origin is shifted are (- 3, 11).

Question 10.

Obtain the new equations of the following loci if the origin is shifted to the point 0′(2,2), the direction of axes remaining the same:

i. 3x-y + 2 = 0

11. x^{2}+y^{2}-3x = 7

iii. xy – 2x – 2y + 4 = 0

iv. y^{2} – 4x – 4y + 12 = 0

Solution:

Given, (h,k) = (2,2)

Let (X, Y) be the new co-ordinates of the point (x,y).

∴ x = X + handy = Y + k

∴ x = X + 2 andy = Y + 2

i. Substituting the values of x and y in the equation 3x -y + 2 = 0, we get

3(X + 2) – (Y + 2) + 2 = 0

∴ 3X + 6-Y-2 + 2 = 0

∴ 3 X – Y + 6 = 0, which is the new equation of locus.

ii. Substituting the values of x and y in the equation

x^{2} + y^{2} – 3x = 7, we get

(X + 2)^{2} + (Y + 2)^{2} – 3(X + 2) = 7

∴ X^{2} + 4X + 4 + Y^{2} + 4Y + 4 – 3X – 6 = 7

∴ X^{2} + Y^{2} + X + 4Y – 5 = 0, which is the new

equation of locus.

iii; Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get

(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0

∴ XY + 2X + 2Y + 4 – 2X – 4-2Y- 4 + 4 = 0

∴ XY = 0, which is the new equation of locus.

iv. Substituting the values of x and y in the equation y^{2} – 4x – 4y + 12 = 0, we get

(Y + 2)^{2} – 4(X + 2) – 4(Y + 2) + 12 = 0

∴ Y^{2} + 4Y + 4 – 4X – 8 – 4Y -8 + 12 = 0

∴ Y^{2} – 4X = 0, which is the new equation of locus.