2x+3y+1=0→‌ perpendicular bisector of ‌AB mAB‌‌=‌
3
2
‌ So, ‌AB:y−2‌‌=‌
3
2
(x−3) ⇒‌‌AB:3x−2y−5‌‌=0 x+2y−12=0→‌ perpendicular bisector of ‌AC mAC‌‌=2 AC:(y−2)‌‌=2(x−3) AC:2x−y−4‌‌=0 On solving 3x−2y−5=0 and 2x+3y+1=0 We get coordinates of D 6x−4y−10=0 6x+9y+3=0 13y+13=0⇒y=−1 and x=1 So, ‌‌D≡(1,−1) and, ‌‌B+A=2D B+(3,2)=(2,−2) ∴‌‌B=(−1,−4) Similarly, on solving Coordinates of E as (4,4) and (3,2)+C‌‌=(8,8)⇒C=(5,6) So, B‌‌=(−1,−4)‌ and ‌C=(5,6) ∴mBC=‌