EAMCET Engineering 2010 Solved Paper

Since, Q is the mid point of PP′, then
Q=(‌

x1+4

2

,‌

y1−13

2

)
Also Q lies on the AB, then
5(‌

x1+4

2

)+(‌

y1−13

2

)+6=0
5x1+20+y1−13+12=0
5x1+y1+19=0 . . . (i)
Since, line PP′ is perpendicular to AB, then (Slope of AB)×(. Slope of PP′)=−1
(−5)×(‌

y1+13

x1−4

)‌=−1
5y1+65‌=x1−4
x1−5y1−69‌=0 . . . (ii)
Solving Eqs. (i) and (ii), we get
25x1+5y1+95=0
x1−5y1−69=0
26x1+26=0
⇒ x1=−1
‌ From Eq. (ii) ‌5y1=−1−69
5y1=−70
y1=−14
So, image of the point p with respect to AB is (−1,−14)