Question

In $△ABC$ the equations of the sides $\overline{AB},\overline{AC}$ are $2x+3y=29,x+2y=16$. lf the mid point of $\overline{BC}$ is $(5,6)$ then the equation of the side $\overline{BC}$ is:

• A. $x+y-11=0$
• B. $2x+y-16=0$
• C. $x-y+1=0$
• D. $2x-y-4=0$

Answer 1

The correct option is A $x+y-11=0$

So, Let $B(h,k)$ then co-ordinates of $C$ are

By mid point formula,

$x=\left(\frac{x_1+x_2}{2}\right)andy=\left(\frac{y_1+y_2}{2}\right)$

$2\times 5=h+x_1$ and $2\times 6=k+y$

$x_1=10-h$ and $y=12-k$

$B$ lies on $2x+3y=29$

So, $2h+3k=29$ ----(1)

And $C$ lie on $x+2y=16$

So, $(10-h)+2(12-k)=16$

$h+2k=18$ ----(2)

From 1 & 2,

$k=7$ and $h=4$

So, equation of line passing through $(4,7)and(5,6)$ is

$y-6=-1(x-5)$

$\Rightarrow y+x=11$