)+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)