
Let the third vertex of △ABC be (a,b). Orthocentre =H(0,0) Let A(5,−1) and B(−2,3) be other two vertices of △ABC. Now, (Slope of AH)×( Slope of BC)=−1 ⇒(5−0−1−0)(a+2b−3)=−1⇒b−3=5(a+2) Similarly, ( Slope of BH)×( Slope of AC)=−1 ⇒−(23)×(a−5b+1)=−1⇒3b+3=2a−10⇒3b−2a+13=0 On solving equation (1) and (2) we get a=−4,b=−7 Hence, third vertex is (−4,−7).