Coordinate Geometry Question 166
Question: 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]
Options:
A) (- 33, -26)
B) (33, 26)
C) (26, 33)
D) None of these
Show Answer
Answer:
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 $
therefore $ 3h-5k+31=0 $ …..(ii)
On solving (i) and (ii) we get $ h=33,k=26 $
Hence the third vertex is (33, 26).
BETA
