In a triangle ABC, side AB has the equation 2 | vedclass

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Question

Let co-ordinates of $B$ be $\left(x_{1}, y_{1}ight) \& C$ be $\left(x_{2}, y_{2}ight)$

$	herefore(5,6)$ is the mid point,

so, $\frac{x_

Vedclass · Accepted Answer

$x + y = 11$

Vedclass · Answer

Let co-ordinates of $B$ be $\left(x_{1}, y_{1}ight) \& C$ be $\left(x_{2}, y_{2}ight)$

$	herefore(5,6)$ is the mid point,

so, $\frac{x_{1}+x_{2}}{2}=5, \frac{y_{1}+y_{2}}{2}=6$

$\Rightarrow x_{1}+x_{2}=10, y_{1}+y_{2}=12$

$B\left(x_{1}, y_{1}ight)$ lies on the line $2 x+3 y=29$

$	herefore 2 x_{1}+3 y_{1}=29---(1)$

$C\left(x_{2}, y_{2}ight)$ lies on the line $x+2 y=16$

$	herefore x_{2}+2 y_{2}=16$

$	herefore$ putting $x_{1}, y_{1}$ in the form of $x_{2}, y_{2}$ in (1)

$2\left(10-x_{2}ight)+3\left(12-y_{2}ight)=29 \quad\left\{x_{1}=10-x_{2}, y_{1}=12-y_{2}ight\}$

$\Rightarrow 20-2 x_{2}+36-3 y_{2}=29$

$\Rightarrow 2 x_{2}+3 y_{2}=27$

on subtracting (3) and $(2) 	imes 2$

$-y_{2}=-5$

$y_{2}=5$

Putting $y_{2}$ in (2)

$x_{2}+2(5)=16$

$x_{2}=6$

$x_{1}=10-x_{2}$

$=4$

$y_{1}=12-5=7$

Equation:

$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$

$\Rightarrow \frac{x-4}{2}=\frac{y-7}{-2}$

$\Rightarrow-x+4=y-7$

$\Rightarrow x+y=11$

In a triangle $ABC,$ side $AB$ has the equation $2 x + 3 y = 29$ and the side $AC$ has the equation , $x + 2 y = 16$ . If the mid - point of $BC$ is $(5, 6)$ then the equation of $BC$ is :

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