The equations of two sides AB and AC of a triangle ABC are 4x+y=14 and 3x−2y=5, respectively. The point (2,−34) divides the third side BC internally in the ratio 2:1. The equation of the side BC is :
(A)x−6y−10=0
(B)x−3y−6=0
(C)x+3y+2=0✓ Correct
(D)x+6y+6=0
Answer: (C)
Solution
Let B(x1,14−4x1) on 4x+y=14 and C(x2,23x2−5) on 3x−2y=5. Since P(2,−34) divides BC in 2:1: 32x2+x1=2 and 32⋅23x2−5+(14−4x1)=−34, giving x1=4,x2=1. So B(4,−2),C(1,−1); slope =−31, line BC:x+3y+2=0. Option (3).
JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option