Question

Find the equations to the straight lines which divide, internally and externally, the line joining $(-3,7)$ to $(5,-4)$ in the ratio of $4:7$ and which are perpendicular to this line.

Answer 1

$A(-3,7),B(5,-4)$

Let $P$ be the point of internal division

$P(\frac{4\left(5\right)+7(-3)}{4+7},\frac{4(-4)+7\left(7\right)}{4+7})P(\frac{-1}{11},\frac{33}{11})$

Slope of $AB$ $=m=\frac{-4-7}{5-(-3)}=-\frac{11}{8}$

Let the slope of line perpendicular to $AB$ be $m^{\prime }$

$m{m}^{\prime }=-1-\frac{11}{8}{m}^{\prime }=-1m^{\prime }=\frac{8}{11}$

Equation of line with slope $m^{\prime }$ and passing through $P$ is

$y-3=\frac{8}{11}(x+\frac{1}{11})y-3=\frac{8}{121}(11x+1)121y-363=88x+8121y-88x=371$

Let $Q$ be the point of external division

$Q(\frac{4\left(5\right)-7(-3)}{4-7},\frac{4(-4)-7\left(7\right)}{4-7})Q(-\frac{41}{3},\frac{65}{3})$

Equation of line with slope $m^{\prime }$ and passing through $Q$ is

$y-\frac{65}{3}=\frac{8}{11}(x+\frac{41}{3})11(3y-65)=8(3x+41)33y-24x=1043$