Question

Find the image of the point (3,8) with respect to the line $x+3y=7$ assuming the line to be a plane mirror.

Answer 1

Let the image of the point A(3,8) in the line mirror DE be $C(\alpha ,\beta )$. Then AC is perpendicular bisector of DE.

The coordinates of pint B are $(\frac{\alpha +3}{2},\frac{\beta +8}{2})$

Since point B lies on the line x+3y=7,

$\therefore \frac{\alpha +3}{2},\frac{3\beta +8}{2}=7$

$\Rightarrow \alpha +3+3\beta +24=14$

$\therefore \alpha +3\beta +13=0$

Since AC is perpendicular on DE,

$\therefore$ slope of AC $\times$ Slope of DE=-1

$\Rightarrow \frac{\beta -8}{\alpha -3}\times \frac{-1}{3}=-1\Rightarrow \beta -8=3\alpha -9$

$\to 3\alpha -\beta -1=0\dots \left(ii\right)$

Solving (i) and (ii) we get

$\alpha =-1$ and $\beta =-4$

Thus image of point (3,8) is (-1, -4)