Question

The image of the point $(3,8)$ with respect to the line $x+3y=7$ is

• A. $(1,4)$
• B. $(4,1)$
• C. $(-1,-4)$
• D. $(-4,-1)$

Answer 1

Answer: (C)

$(-1,-4)$

Let $Q(x_1,y_1)$ be the image of the point $P(3,8)$ in the line $x+3y=7$.
Then, $PQ$ is perpendicular to the given line. So,
$\frac{y_1-8}{x_1-3}x-\frac{1}{3}=-1$
$\Rightarrow 3x_1-y_1=1....\left(i\right)$
Also, the mid-point of $PQ$ i.e., $R(\frac{x_1+3}{2},\frac{y_1+8}{2})$ lies on the line $x+3y=7$.
$\therefore \frac{x_1+3}{2}+3\left(\frac{y_1+8}{2}\right)=7$
$\Rightarrow x_1+3y_1+3+24=14$
$\Rightarrow x_1+3y_1+13=0....\left(ii\right)$
On solving Eqs. (i) and (ii), we get
$Q(-1,-4)$.