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 ⇒(
−1−0
5−0
)(
b−3
a+2
)=−1 ⇒b−3=5(a+2) . . . (i) Similarly, ( Slope of BH)×( Slope of AC)=−1 ⇒−(
3
2
)×(
b+1
a−5
)=−1 ⇒3b+3=2a−10 ⇒3b−2a+13=0 . . . (ii) On solving equations (i) and (ii) we get a=−4,b=−7 Hence, third vertex is (-4,-7)