# Q : 18    Find the image of the point  $\small (3,8)$  with respect to the line  $x+3y=7$  assuming the  line to be a plane mirror.

Answers (1)
G Gautam harsolia

Let point $(a,b)$ is the image of point $\small (3,8)$ w.r.t. to line $x+3y=7$
line $x+3y=7$  is perpendicular bisector of line joining points  $\small (3,8)$  and $(a,b)$
Slope of line $x+3y=7$ , $m' = -\frac{1}{3}$
Slope of   line joining points  $\small (3,8)$  and $(a,b)$  is  , $m = \frac{8-b}{3-a}$
Now,
$m = -\frac{1}{m'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because lines \ are \ perpendicular)$
$\frac{8-b}{3-a}= 3$
$8-b=9-3a$
$3a-b=1 \ \ \ \ \ \ \ \ \ \ \ -(i)$
Point of intersection is the midpoint of line  joining points  $\small (3,8)$  and $(a,b)$
Therefore,
Point of intersection is  $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$
Point $\left ( \frac{3+a}{2},\frac{b+8}{2} \right )$ also  satisfy the line  $x+3y=7$
Therefore,
$\frac{3+a}{2}+3.\frac{b+8}{2}=7$
$a+3b=-13 \ \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
On solving equation (i) and (ii) we will get
$(a,b) = (-1,-4)$
Therefore, the image of the point  $\small (3,8)$  with respect to the line  $x+3y=7$ is $(-1,-4)$

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