Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2 and x + y = 2, respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to
- 8
- 4
- 10
- 6
Answer
(4) 6
Solution
B: put y = 0 in 3y - x = 2 B(−2, 0)
C: put y = 0 in x + y = 2 C(2, 0)
A: solving 3y - x = 2 and x + y = 2 4y = 4 A(1, 1)
Altitude from A ⟂ BC (x-axis): x = 1
Altitude from B ⟂ AC (slope −1): slope 1 y = x + 2
orthocentre P(1, 3)
area of PBC = 12 BC 3 = 12 4 3 = 6
Answer: (4)