Question

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

Answer 1

Let image of a point be $B(h,k)$. Mid point of $AB$ is $B(\frac{h-8}{2},\frac{k+12}{2})$ which lies on a line
$4x+7y+13=0$
$\therefore$ $4\left(\frac{h-8}{2}\right)+7\left(\frac{k+12}{2}\right)+13=0$
$\Rightarrow$ $4h+7k+78=\mathrm{0......}\left(i\right)$
Slope of given line is $-\frac{4}{7}$ then slope of line $AB$ is $\frac{7}{4}$
$\therefore$ $\frac{k-12}{h+8}=\frac{7}{4}$
$\Rightarrow$ $4k-7h-104=\mathrm{0.......}\left(ii\right)$
On solving eqs, $\left(i\right)$ and $\left(ii\right)$ we get
$k=-2$ and $h=-16$
Hence required image is $(-16,-2)$