**Chapter 5 Straight Line Ex 5.1**

## 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

-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).

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

**Question 4.**

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

∴ 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,

**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 pointsi. 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 pointsi. 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 = 011. x**

^{2}+y

^{2}-3x = 7iii. xy – 2x – 2y + 4 = 0iv. y

^{2}– 4x – 4y + 12 = 0Solution:

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

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.