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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 :
$x - y = - 1$
$5 x - 2 y = 13$
$x + y = 11$
$3 x - 4 y = - 9$
Solution
Let co-ordinates of $B$ be $\left(x_{1}, y_{1}\right) \& C$ be $\left(x_{2}, y_{2}\right)$
$\therefore(5,6)$ is the mid point,
so, $\frac{x_{1}+x_{2}}{2}=5, \frac{y_{1}+y_{2}}{2}=6$
$\Rightarrow x_{1}+x_{2}=10, y_{1}+y_{2}=12$
$B\left(x_{1}, y_{1}\right)$ lies on the line $2 x+3 y=29$
$\therefore 2 x_{1}+3 y_{1}=29—(1)$
$C\left(x_{2}, y_{2}\right)$ lies on the line $x+2 y=16$
$\therefore x_{2}+2 y_{2}=16$
$\therefore$ putting $x_{1}, y_{1}$ in the form of $x_{2}, y_{2}$ in (1)
$2\left(10-x_{2}\right)+3\left(12-y_{2}\right)=29 \quad\left\{x_{1}=10-x_{2}, y_{1}=12-y_{2}\right\}$
$\Rightarrow 20-2 x_{2}+36-3 y_{2}=29$
$\Rightarrow 2 x_{2}+3 y_{2}=27$
on subtracting (3) and $(2) \times 2$
$-y_{2}=-5$
$y_{2}=5$
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:
$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}$
$\Rightarrow \frac{x-4}{2}=\frac{y-7}{-2}$
$\Rightarrow-x+4=y-7$
$\Rightarrow x+y=11$