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 A (3,8) be B (a,b). Also, let M be the midpoint of AB.

$\therefore \mathrm{Coordinates}\mathrm{of}M=\left(\frac{3+a}{2},\frac{8+b}{2}\right)$



Point M lies on the line x + 3y = 7

$\therefore \frac{3+a}{2}+3\times \left(\frac{8+b}{2}\right)=7$

$\Rightarrow a+3b+13=0$ ... (1)

Lines CD and AB are perpendicular.

∴ Slope of AB $\times$ Slope of CD = −1

$$\begin{gathered} \Rightarrow \frac{b-8}{a-3}\times \left(-\frac{1}{3}\right)=-1 \\ \Rightarrow b-8=3a-9 \end{gathered}$$

$\Rightarrow 3a-b-1=0$ ... (2)

Solving (1) and (2) by cross multiplication, we get:

$$\begin{gathered} \frac{a}{-3+13}=\frac{b}{39+1}=\frac{1}{-1-9} \\ \Rightarrow a=-1,b=-4 \end{gathered}$$

Hence, the image of the point (3, 8) with respect to the line mirror x + 3y = 7 is (−1, −4).