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)

$\left(\frac{6}{5}, \frac{7}{5}\right)$

Answer 2

Let $Q(h, k)$ be the image of the point $P(1, 2)$ in the line $x - 3y + 4 = 0\quad \ldots (i)$ 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 $3h+k=5 \quad\ldots\left(ii\right)$ 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 $h-3k=-3 \quad\ldots\left(iii\right)$ 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 $\left(\frac{6}{5}, \frac{7}{5}\right)$.