Question

The vertices of triangle $ABC$ are $A(1,-2),B(-7,6)$ and $C(11/5,2/5)$.

Answer 1

A) Let mid point of $A(1,-2)$ and $B(-7,6)$ is $D=(\frac{1-7}{2},\frac{-2+6}{2})=(-3,2)$
Slope of $m\left(AB\right)=\frac{6+2}{-7-1}=-1$ and slope of perpendicular line is $m=1$
Hence equation of line with $m=1$ and passing through point D is
$\frac{y-2}{x+3}=1\Rightarrow y-2=x+3\Rightarrow x-y+5=0$
B) Let mid point of $B(-7,6)$ and $C(\frac{11}{5},\frac{2}{5})$ is $E=(\frac{-7+\frac{11}{5}}{2},\frac{6+\frac{2}{5}}{2})=(-\frac{12}{5},\frac{16}{5})$
Hence equation of line through $A(1,-2)$ and $E(-\frac{12}{5},\frac{16}{5})$ is
$\frac{y+2}{x-1}=\frac{\frac{16}{5}+2}{-\frac{12}{5}-1}\Rightarrow 26x+17y+8=0$
C) Slope of $m\left(AB\right)=-1$, then slope of perpendicular line is $m=1$
Hence equation of line passing through $C(\frac{11}{5},\frac{2}{5})$ and slope $m=1$ is
$\frac{y-\frac{2}{5}}{x-\frac{11}{5}}=1\Rightarrow y-\frac{2}{5}x-\frac{11}{5}\Rightarrow 5x-5y-9=0$
D) Equation of line passing through $B(-7,6)$ and $C(\frac{11}{5},\frac{2}{5})$ is
$\frac{y-\frac{2}{5}}{6-\frac{2}{5}}=\frac{x-\frac{11}{5}}{-7-\frac{11}{5}}\Rightarrow \frac{5y-2}{28}=\frac{5x-11}{-46}\Rightarrow 41x+23y-40=0$