Question

The point P(2, 4) is first reflected on the line y = x and then the image point Q is again reflected on the line y = – x to get the image point Q'. Then the circumcentre of the ΔPQO' is

Options

• (0, 0)

• (– 2, – 4)

• (4, 2)

• (4, – 2) 

Answer 1

(0, 0)

Explanation:

Given point P : (2, 4)

Hence, reflection of point p in y = x is (4, 2)

Let Q : (4, 2)

While reflection of Q in y = – x is (2, – 4)

Let Q' (2, – 4)

Let circumcentre of the ΔPQQ' is O : (x, y)

Hence OP = OQ = OQ'

For OP ⇒ (x, y) (2, 4)

OP = `sqrt((4 - y)^2 + (2 - x)^2`

OQ = `sqrt((2 - y)^2 + (4 - x)^2`

OQ' = `sqrt((-4 - y)^2 + (2 - x)^2`

OP² = OQ² ⇒ (4 – y)² + (2 – x)² = (2 – y)² + (4 – x)²

OQ² = OQ′² ⇒ (2 – y)² + (4 – x)² = (– 4 – x)² + (2 – k)² 

⇒ x = 0 and y = 0

Hence, O : (0, 0)