Equation of the perpendicular bisectors of AB:x−y=0 AC:x+y=2 So, mAB=+1 and mAC=−1 Equation of AB⇒y−3=+1(x−2) x−y=−1 Equation of AC⇒y−3=−1(x−2) ⇒x+y=5
A+C=2D and A+B=2E (2,3)+(C)=(1,3) and (2,3)+(B)=(5,5) C=(−1,0) and B=(3,2) So, equation of BC:y=