Coordinate Geometry Question 168
Question: Orthocentre of the triangle whose vertices are (0, 0) (2, -1) and (1, 3) is
[ISM Dhanbad1970; IIT 1967, 74]
Options:
A) $ ( \frac{4}{7},\frac{1}{7} ) $
B) $ ( -\frac{4}{7},-\frac{1}{7} ) $
C) (-4, -1)
D) (4, 1)
Show Answer
Answer:
Correct Answer: B
Solution:
Equation of line BC is $ 4x+y=7 $ , then equation of line AD is $ x-4y+k=0 $ but it passes through (0,0) hence $ k=0 $ thus equation of $ AD=x-4y=0 $ …..(i) Similarly the equation of $ AC=3x-y=0 $ and BE is $ x+3y+1=0 $ …..(ii)
On solving (i) and (ii), the required orthocentre is $ ( \frac{-4}{7},\frac{-1}{7} ) $ .
BETA
