Question

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

• A. $(-2,-1)$
• B. $(-1,-2)$
• C. $\left(\frac{6}{5},\frac{7}{5}\right)$
• D. $\left(\frac{-6}{5},\frac{-7}{5}\right)$

Answer 1

Answer: (C)

Let $Q(h,k)$ be the image of the point $P(1,2)$ in the line $x-3y+4=0\dots (i)$
Therefore, the line $(i)$ is the perpendicular bisector of line segment $PQ$.
Hence, slope of line $PQ=\frac{-1}{\text{Slope of line x - 3y + 4 = 0}}$
so that $\frac{k-2}{h-1}=\frac{-1}{\frac{1}{3}}$ or $3h+k=5\dots \left(ii\right)$
and the mid-point of $PQ$, i.e., point $\left(\frac{h+1}{2},\frac{k+2}{2}\right)$ will satisfy the equation $\left(i\right)$ so that
$\frac{h+1}{2}-3\left(\frac{k+2}{2}\right)+4=0$ or $h-3k=-3\dots \left(iii\right)$
Solving $\left(ii\right)$ and $\left(iii\right)$,
we get $h=\frac{6}{5}$ and $k=\frac{7}{5}$.
Hence, the image of the point $\left(1,2\right)$ in the line $\left(i\right)$ is
$\left(\frac{6}{5},\frac{7}{5}\right)$.