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Question: Two vertices of a triangle are (4, –3) and (–2, 5). If the orthocentre of the triangle is at (1, 2),...

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.

A

(– 33, – 26)

B

(33, 26)

C

(26, 33)

D

None of these

Answer

(33, 26)

Explanation

Solution

Let third vertex be (h, k). Now slope of AD is k−2h−1\frac { k - 2 } { h - 1 }

Slope of BCB C is 5+3−2−4=−43\frac { 5 + 3 } { - 2 - 4 } = \frac { - 4 } { 3 }

Slope of BE is −3−24−1=−53\frac { - 3 - 2 } { 4 - 1 } = \frac { - 5 } { 3 }

And slope of AC is k−5h+2\frac { k - 5 } { h + 2 }

Since AD⊥BC⇒k−2h−1×−43=−1A D \perp B C \Rightarrow \frac { k - 2 } { h - 1 } \times \frac { - 4 } { 3 } = - 1

3h−4k+5=03 h - 4 k + 5 = 0 ......(i)

Again Since BE⊥AC⇒−53×k−5h+2=−1B E \perp A C \Rightarrow - \frac { 5 } { 3 } \times \frac { k - 5 } { h + 2 } = - 1

⇒ 3h−5k+31=03 h - 5 k + 31 = 0 .....(ii)

on solving (i) and (ii) we get h=33,k=26h = 33 , k = 26

Hence the third vertex is (33, 26).