Straight Lines
Triangle — Equation of Third Side from Internal Division
nta_pyq_2024_apr
Grade 11
Question:
The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x+y=14$ and $3x-2y=5$, respectively. The point $\left(2,-\dfrac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is
$x+3y+2=0$
$x-6y-10=0$
$x-3y-6=0$
$x+6y+6=0$
Step-by-Step Solution
Key Concept: Find $B$ on $AB$: $B=(x_1,14-4x_1)$. Find $C$ on $AC$: $C=(x_2,(3x_2-5)/2)$. Internal division $2:1$: $\frac{2x_2+x_1}{3}=2$ and $\frac{2\cdot\frac{3x_2-5}{2}+(14-4x_1)}{3}=-\frac{4}{3}$.
$B=(4,-2)$, $C=(1,-1)$. Equation of $BC$: $x+3y+2=0$.
Correct Answer: 1