Question Number 38880 by Rio Mike last updated on 30/Jun/18
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{perpendicular} \\ $$$${bisector}\:{of}\:{the}\:{line}\:{segment}\:{joining} \\ $$$${the}\:{points}\:\left(\mathrm{1},\mathrm{3}\right)\:{and}\:\left(\mathrm{5},\mathrm{1}\right) \\ $$
Answered by MrW3 last updated on 30/Jun/18
$${eqn}\:{of}\:{line}\:{joining}\:\left(\mathrm{1},\mathrm{3}\right)\:{and}\:\left(\mathrm{5},\mathrm{1}\right): \\ $$$$\frac{{y}−\mathrm{1}}{\mathrm{3}−\mathrm{1}}=\frac{{x}−\mathrm{5}}{\mathrm{1}−\mathrm{5}} \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{5}\right)+\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$ \\ $$$${midpoint}\:{of}\:\left(\mathrm{1},\mathrm{3}\right)\:{and}\:\left(\mathrm{5},\mathrm{1}\right): \\ $$$$\left(\mathrm{3},\mathrm{2}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{perpendicular}\:{bisector}: \\ $$$$\frac{{y}−\mathrm{2}}{{x}−\mathrm{3}}=−\frac{\mathrm{1}}{−\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{2} \\ $$$$\Rightarrow{y}=\mathrm{2}{x}−\mathrm{4} \\ $$