Question

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

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

Answer 1

The correct option is C $(-1,-4)$

Let the image be $(x_1,y_1)$
$\therefore (\frac{3+x_1}{2},\frac{8+y_1}{2})$ will be lie on the given line.
$\therefore \frac{3+x_1}{2}+\frac{3}{2}(8+y_1)=7$
$\Rightarrow x_1+3y_1+13=0...\left(i\right)$
Now line joining $(x_1,y_1)$ with $(3,8)$ in perpendicular to the given line
$\therefore \left(\frac{y_1-8}{x_1-3}\right)\times (-\frac{1}{3})=-1$
$y_1-8=3(x_1-3)$
$3x_1-y_1=1...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$(x_1,y_1)=(-1,-4)$

Alternate Solution:
Using the image formula
$\frac{x-3}{1}=\frac{y-8}{3}=-\frac{2(3+3\cdot 8-7)}{1^2+3^2}\Rightarrow x=-1;y=-4$
So image will be
$(-1,-4)$