Question

Assuming that straight lines work as the plane mirror for a point, find the image of the point $(1,2)$ in the line $x-3y+4=0$.

Answer 1

let $(h,k)$ be image

$(x_1,y_1)=(1,2)$ is given point

comparing $x-3y+4=0$ with

$ax+by+c=0$

We have ,

$a=1b=-3c=4$

By image theorem

$\frac{h-x_1}{a}=\frac{k-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2}$

$\Rightarrow \frac{h-1}{1}=\frac{k-2}{-3}=\frac{-2(\left(1\right)\left(1\right)+(-3)\left(2\right)+4)}{{\left(1\right)}^2+{(-3)}^2}$

$\Rightarrow h-1=\frac{k-2}{-3}=\frac{-2(1-)}{10}$

$\Rightarrow h-1=\frac{2}{10}\frac{k-2}{-3}=\frac{2}{10}$

$\Rightarrow h=\frac{6}{5}k=\frac{7}{5}$

Image $=(h,k)=(\frac{6}{5},\frac{7}{5})$