Equations of two sides mathrm{AB} and mathrm{AC} of a triangle mathrm{ABC} are 4 mathrm{x}+mathrm{y}

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

A

$x-6 y-10=0$

B

$x-3 y-6=0$

C

$x+3 y+2=0$

D

$x+6 y+6=0$

(JEE MAIN-2024)

Solution

$ \frac{2 x_2+x_1}{3}=2, \frac{2\left(\frac{3 x_2-5}{2}\right)+\left(14-4 x_1\right)}{3}=\frac{-4}{3} $

$ 2 x_2+x_1=6,3 x_2-4 x_1=-13 $

$ x_2=1, x_1=4$

So, $\mathrm{C}(1,-1), \mathrm{B}(4,-2)$

$\mathrm{m}=\frac{-1}{3}$

Equation of $B C: y+1=\frac{-1}{3}(x-1)$

$ 3 y+3=-x+1$

$ x+3 y+2=0$

Standard 11

Mathematics

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