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

Equation of the given line is
4x + 7y + 13 = 0                                                                    (1)
Let Q (a, b) be the image of the point P(–8, 12) w.r.t. line (1).
Then, PQ $\perp$ line (1) and PC = CQ.
Equation of the line PC is
$\left(y-12\right)=\frac{7}{4}(x+8)$
[PC is$\perp$to the line (1) and passes through (–8, 12)]
or 7x – 4y + 104 = 0                                                                (2)
Solving Eq. (1) and (2), we get
x = –12 and y = 5.          $\therefore$C ≡ (–12, 5)

Since C is mid-point of PQ,

$\therefore -12=\frac{\alpha -8}{2} and 5=\frac{\beta +12}{2}$

$\Rightarrow \alpha =-16 and \beta =-2$

$\therefore Q=(-16,-2)$