AP EAMCET Engineering 2011 Solved Paper

The equation of perpendicular line on x+3y=7 is 3x−y+λ=0.
Since, it is passes through (3,8).
∴ 9−8+λ=0
⇒λ=−1
∴ 3x−y−1=0
The point of intersection of lines x+3y=7 and 3x−y−1=0 is (1,2), which is the foot of a point.
Let the image of a point (3,8) be (x1,y1).
∴

3+x1

2

=1,

8+y1

2

=2
⇒ x1=−1,y1=−4