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

Ch. 10
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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