Two vertices of a triangle are (4,−3) and(−2,5) . If the ortho centre of the triangle is at (1,2) , then the third vertex is

Let the third vertex be (h, k)Now slope of AD is (k−2)(h−1) slope of BC is (5+3)(−2−4)=−4/3 ,slope of BE is(−3−2)(4−1)=53 And slope of AC is (k−5)(h+2) .Since AD⊥BC , sok−2h−1×−43=−1⇒3h−4k+5=0 ……………(1)Again since BE⊥AC , so−53×k−5h+2=−1⇒3h−5k+31=0 ……………(2)On solving (1) and (2) we geth=33,k=26 . Hence the third vertex is (33,26)