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

	Column-I		Column-II
A)	Equation of the perpendicular bisector of AB	P)	$26\mathrm{x}+17\mathrm{y}+8=0$
B)	Equation of the median through A	Q)	$14\mathrm{x}+23\mathrm{y}-40=0$
C)	Equation of the altitude through C	R)	$\mathrm{x}-\mathrm{y}+5=0$
D)	Equation of BC	S)	$5\mathrm{x}-5\mathrm{y}-9=0$

	(A)	(B)	(C)	(D)
1)	Q	P	S	R
2)	R	P	S	Q
3)	R	Q	S	P
4)	S	Q	R	P

 

Equation of line is $\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

$\text{ Let mid point of }\mathrm{A}(1,-2)\text{ and }\mathrm{B}(-7,6)\text{ is }\mathrm{D}=\left(\frac{1}{2},\frac{1-0}{2}\right)=(-3,2)$

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

$$\begin{gathered} \text{ Hence equation of line with }m=1\text{ and passing through point }D\text{ is } \\ x-y+5=0 \end{gathered}$$

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

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

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

$\text{ Slope of }m\left(AB\right)=-1\text{, then slope of perpendicular line is }m=1$

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

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

$\text{ Equation of line passing through }\mathrm{B}(-7,6)\text{ and }\mathrm{C}\left(\frac{11}{5},\frac{2}{5}\right)\text{ is }$

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