question_answer

Two vertices of a triangle are (4, -3) and (-2, 5). If the orthocentre of the triangle is at (1, 2), then the third vertex is  [Roorkee 1987]

A) (- 33, -26)

B) (33, 26)

C) (26, 33)

D) None of these

Correct Answer: B

Solution :

Let third vertex be (h, k). Now slope of AD is \[\frac{k-2}{h-1}\], Slope of \[BC\] is \[\frac{5+3}{-2-4}=\frac{-4}{3}\] Slope of BE  is \[\frac{-3-2}{4-1}=\frac{-5}{3}\] And slope of AC is \[\frac{k-5}{h+2}\]  Since  \[AD\,\bot \,BC\Rightarrow \frac{k-2}{h-1}\times \frac{-4}{3}=-1\]   \[3h-4k+5=0\]  ......(i) Again Since \[BE\,\bot \,AC\Rightarrow -\frac{5}{3}\times \frac{k-5}{h+2}=-1\] Þ \[3h-5k+31=0\]  .....(ii) on solving (i) and (ii) we get \[h=33,k=26\] Hence the third vertex is (33, 26).