Question

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point \( \left(2,-\frac{4}{3}\right)\) divides the third side BC internally in the ratio 2 : 1. The equation of the side BC is :

(1) x – 6y – 10 = 0 

(2) x – 3y – 6 = 0 

(3) x + 3y + 2 = 0 

(4) x + 6y + 6 = 0

Answer 1

Correct option is (3) x + 3y + 2 = 0   

\(\frac{2 \alpha+\beta}{3}=2\)

\(2 \alpha+\beta=6\)                     .......(1)

\(\frac{3 \alpha-5+14-4 \beta}{3}=\frac{-4}{3}\)

\(3 \alpha-4 \beta=-13 \quad \)         .......(2)

\(4(2 \alpha+\beta=6)\)

\(3 \alpha-4 \beta=-13\)

\(11 \alpha=11\)

α = 1

\(2+\beta=6\)

\(\beta=4\)

Point C(1, –1 )

Point B(4, –2)

eq. of BC

\(y+1 = \frac{-2+1}{4-1}\,(x-1)\)

3y + 3 = -(x - 1)

3y + x + 2 = 0