Matrix Matching :

The vertices of triangle $ABC$ are $A(1,-2),B(-7,6)$ and $C(\frac{11}{5},\frac{2}{5})$

Given  $A(1,-2)B(-7,6)C=(\frac{11}{5},\frac{2}{5})$

Option (A) :  Midpoint of  $\overline{AB}=(\frac{1-7}{2},\frac{-2+6}{2})=(-3,2)$

Slope of $\overleftrightarrow{AB}=\frac{6+2}{-7-1}=\frac{8}{-8}=-1$

Equation of  ${\perp }^{lar}$ bisector  $\overline{AB}$ is

$y-y_1=m(x-x_1)$

$\Rightarrow y-2=1(x+3)$

$\Rightarrow y-2=x+3$

$x-y+5=0$

Option (B) : Midpoint of  $\overline{BC}=(\frac{1}{2}(-7+\frac{11}{5}),\frac{1}{2}(6+\frac{2}{5}))$

$D=(\frac{-24}{2\times 5},\frac{32}{2\times 5})$

$=(\frac{-12}{5},\frac{16}{5})$

Equation of  $AD$ is $=(y+2)(\frac{-12}{5}-1)=(x-1)(\frac{-16}{5}+2)$

$\Rightarrow (y+2)\left(\frac{-17}{5}\right)=(x-1)\left(\frac{26}{5}\right)$

$\Rightarrow -17y-34=26x-26$

$26x+17y+8=0$

Option ( C) : $C=(\frac{11}{5},\frac{2}{5})$

Slope of $\overline{AB}=-1$

Slope of $\overline{CM}=1$

Equation of  $\overline{CM}$ is

$y-y_1=m(x-x_1)$

$\Rightarrow y-\frac{2}{5}=1(x-\frac{11}{5})$

$\Rightarrow \frac{5y-2}{5}=\frac{5x-11}{5}$

$\Rightarrow 5y-2=5x-11$

$5x-5y-9=0$

Option (D ) : Equation of $\overline{BC}$is  $(y-6)(\frac{11}{5}+7)=(x+7)(\frac{2}{5}-6)$

$(y-6)\left(\frac{46}{5}\right)=(x+7)\left(\frac{2-30}{5}\right)$

$46y-46\times 6=-28x-28\times 7$

$28x+46y-276+196=0$

$28x+46y-80=0$

$14x+23y-40=0$