Question

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

• A. $2x+y=7$
• B. $x+y=1$
• C. $2x-y=17$
• D. None of these

Answer 1

The correct option is B $x+y=1$

Let co-ordinates of $B$ be $(x_1,y_1)$ & $C$ be $(x_2,y_2)$

$\therefore$ $(5,6)$ is the mid point,

so, $\frac{x_1+x_2}{2}=5,\frac{y_1+y_2}{2}=6$

$\Rightarrow x_1+x_2=10,y_1+y_2=12$

$B(x_1,y_1)$ lies on the line $2x+3y=29$

$\therefore 2x_1+3y_1=29$ ----(1)

$C(x_2,y_2)$ lies on the line $x+2y=16,$

$\therefore x_2+2y_2=16$ ----(2)

$\therefore$ putting $x_1,y_1$ in the form of $x_2,y_2$ in (1)

$2(10-x_2)+3(12-y_2)=29$ {$x_1=10-x_2,y_1=12-y_2$}

$\Rightarrow 20-2x_2+36-3y_2=29$

$\Rightarrow 2x_2+3y_2=27$ ----(3)

on subtracting $\left(3\right)$ and $\left(2\right)$ $\times$ $2$

$-y_2=-5$

$y_2=5$

Putting $y_{2}in\left(2\right)$

$x_2+2\left(5\right)=16$

$x_2=6$

$x_1=10-x_2$

$=4$

$y_1=12-5=7$

Equation :

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$

$\Rightarrow \frac{x-4}{2}=\frac{y-7}{-2}$

$\Rightarrow -x+4=y-7$

$\Rightarrow x+y=11$