Find the image of the point $(-8,12)$ with respect to the line mirror $4x+7y+13=0$.

The correct option is C $(-16,-2$)

$M=(\frac{a-8}{2},\frac{b+12}{2})$ is the mid point of AA'.
It lies on $4x+7y+13=0$
So, $\Rightarrow 4\left(\frac{a-8}{2}\right)+7\left(\frac{b+12}{2}\right)+13=0$
$\Rightarrow 2(a-8)+\frac{7b+84}{2}+13=0$
$\Rightarrow 4a-32+7b+84+26=0$
$\therefore 4a+7b+78=0$......()

Since AA' $\perp$ PQ, so
Slope (PQ) $\times$ Slope (AA') = -1
$\frac{-4}{7}\times \frac{b-12}{a+8}=-1$
$-4b+48=-7a-56$
$7a-4b+104=0$......(2)
Multiplying equations (1) by 4 and equation (2) by 7, we get
$16a+28b=-312$
$49a-28b=-728$
By adding above two equations, we get
$16a+49a=-312-728$
$65a=-1040$
$\therefore a=-16$
Substitute $a=-16$ in equation (1), we get
$\Rightarrow 4(-16)+7b+78=0$
$-64+7b+78=0$
$7b+14=0$
$7b=-14$
$\therefore b=-2$
Hence, required image point is $(-16,-2)$.