Question

Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror.

Answer 1

Let the image of P(3, 8) and P’(a, b)

Let the point of intersection be O

Slope of x + 3y = 7 is `-1/3`

Slope of PP’ = 3  ...(perpendicular)

Equation of PP’ is

y – y₁ = m(x – x₁)

y – 8 = 3(x – 3)

y – 8 = 3x – 9

– 8 + 9 = 3x – y

∴ 3x – y = 1   ...(1)

The two line meet at 0

                                    x + 3y = 7       ...(2)
(1) × 3 ⇒                   9x – 3y = 3        ...(3)
(2) × 1 ⇒                    x + 3y = 7        ...(4)
Adding (1) and (2) ⇒ 10x     = 10

x = `10/10` = 1

Substitute the value of x = 1 in (1)

3 – y = 1

3 – 1 = y

2 = y

The point O is (1, 2)

Mid point of pp’ = `((x_1 + x_2)/2, (y_1 + y_2)/2)`

(1, 2) = `((3 + "a")/2, (8 + "b")/2)`

∴ `(3 + "a")/2` = 1

⇒ 3 + a = 2

a = 2 – 3 = – 1

`(8 + "b")/2` = 2

8 + b = 4

b = 4 – 8 = – 4

The point P’ is (– 1, – 4)