Question

The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is

• A. (16, –2)
• B. (–16, 2)
• C. (16, 2)
• D. (–16, –2)

Answer 1

The correct option is D

(–16, –2)

Equation of the given line is

4x + 7y + 13 = 0 …… (1)

Let $Q(\alpha ,\beta )$ be the image of the point P(–8, 12) with respect to line (1)

Then, PQ $\perp$ line (1) and PC = CQ

Equation of the line PC is

$(y-12)=\frac{7}{4}(x+8)$

[PC is $\perp$ to the line (1) and passes through (–8, 12)]

or 7x – 4y + 104 = 0 …… (2)

solving (1) and (2), we get

x = –12 and y = 5.

$\therefore C\equiv (-12,5)$

Since C is mid point of PQ,

$\therefore -12=\frac{\alpha -8}{2}and5=\frac{\beta +12}{2}$

$\Rightarrow \alpha =-16and\beta =-2\therefore Q\equiv (-16,-2)$