Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and 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) and B(2, 1).

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} + 3y^{2} + 4x – 24y + 32 = 0

∴ The required equation of locus 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 if PA^{2} = 3PB^{2}.

Solution:

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

Given, A(4, 1), B(5, 4) and PA^{2} = 3PB^{2}

∴ (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

∴ 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 a right angle at P. (∠APB = 90°)

Solution:

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

Given, A(1, 6) and B(3, 5), ∠APB = 90°

∴ ΔAPB is a right-angled triangle.

By Pythagoras theorem,

AP^{2} + PB^{2} = AB^{2}

∴ [(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

Question 7.

If the origin is shifted to the point O'(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points (a) A(1, 3) (b) B(2, 5)

Solution:

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

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

∴ x = X + h and y = Y + k

∴ x = X + 2 and y = Y + 3 …..(i)

(a) Given, A(x, y) = A(1, 3)

x = X + 2 and y = 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).

(b) Given, B(x, y) = B(2, 5)

x = X + 2 andy = 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 (a) C(5, 4) (b) D(3, 3)

Solution:

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

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

x = X + h and y = Y + k

∴ x = X + 1 and 7 = Y + 3 …..(i)

(a) Given, C(X, Y) = C(5, 4)

∴ x = X + 1 andy = Y + 3 …..[From(i)]

∴ x = 5 + 1 = 6 and y = 4 + 3 = 7

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

(b) Given, D(X, Y) = D(3, 3)

∴ x = X + 1 and y = 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 (5, 14) change to (8, 3) by the 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, (x,y) = (5, 14), (X, Y) = (8, 3)

Since, x = X + h and y = Y + k

∴ 5 = 8 + h and 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 O'(2, 2), the direction of axes remaining the same:

(a) 3x – y + 2 = 0

(b) x^{2} + y^{2} – 3x = 7

(c) xy – 2x – 2y + 4 = 0

Solution:

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

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

∴ x = X + h and y = Y + k

∴ x = X + 2 and y = Y + 2

(a) 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

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

(b) 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.

(c) 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.