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

Column-I	Column-II
A) Equation of the perpendicular  bisector of AB	P) $26x+17y+8=0$
B) Equation of the median through A	q) $14x+23y-40=0$
C) Equation of altitude through C	r) $x-y+5=0$
D) Equation of BC	s) $5x-5y-9=0$

A)$\text{ mid point of }\mathrm{A}(1,-2)\text{ and }\mathrm{B}(-7,6)\text{ is }\mathrm{D}=\left(\frac{1-7}{2},\frac{-2+6}{2}\right)=(-3,2)$

$\text{ Slope of }m\left(AB\right)=\frac{6+2}{-7-1}=-1\text{ and slope of perpendicular line is }m=1$

Equation of perpendicular bisector of AB is

$\frac{y-2}{x+3}=1\Rightarrow y-2=x+3\Rightarrow \boxed{x-y+5=0}$

B)$\text{ mid point of }B(-7,6)\text{ and }C\left(\frac{11}{5},\frac{2}{5}\right)\text{ is }E=\left(\frac{-7+\frac{11}{5}}{2},\frac{6+\frac{2}{5}}{2}\right)=\left(-\frac{12}{5},\frac{16}{5}\right)$

$\text{ Hence equation of}\text{tmedian hrough }\mathrm{A}(1,-2)\text{ and }\mathrm{E}\left(-\frac{12}{5},\frac{16}{5}\right)\text{ is }$

$\frac{y+2}{x-1}=\frac{\frac{16}{5}+2}{-\frac{12}{5}-1}\Rightarrow \boxed{26x+17y+8=0}$

C)$\text{slope of }\mathrm{m}\left(\mathrm{AB}\right)=-1\text{, then slope of altitude is }\mathrm{m}=1$

$\text{ equation of altitude passing through }\mathrm{C}\left(\frac{11}{5},\frac{2}{5}\right)\text{ and slope }\mathrm{m}=1\text{ is }$

$\frac{y-\frac{2}{5}}{x-\frac{11}{5}}=1\Rightarrow y-\frac{2}{5}x-\frac{11}{5}\Rightarrow \boxed{5x-5y-9=0}$

D)$B(-7,6)\text{ and }C\left(\frac{11}{5},\frac{2}{5}\right)$

Equation of BC is

$\frac{y-\frac{2}{5}}{6-\frac{2}{5}}=\frac{x-\frac{11}{5}}{-7-\frac{11}{5}}\Rightarrow \frac{5y-2}{28}=\frac{5x-11}{-46}\Rightarrow \boxed{14x+23y-40=0}$