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

Let the third vertex be (h,k) Now, the slope of AD is (k−2)/(h−1), the slope of BC is (5+3)/(−2−4)=−4/3,the slope of BE is (−3−2)/(4−1)=−5/3, and the slope of AC is (k−5)/(h+2). Since AD⊥BC, we have k−2h−1×−43=−1or 3h−5k+31=0_______(i)Again, since BE⊥AC,, we have −53×k−5h+2=−1or 3h−5k+31=0______(ii)On solving (i) and (ii), we get h=33, k=26. Hence, the third vertex is(33,26)