Question Number 137412 by oustmuchiya@gmail.com last updated on 02/Apr/21
Answered by herbert last updated on 02/Apr/21
$${gradient}\:{of}\:{l}_{\mathrm{1}} \:=\:\frac{\mathrm{2}+\mathrm{4}}{\mathrm{5}+\mathrm{1}}\:=\frac{\mathrm{6}}{\mathrm{6}}=\mathrm{1} \\ $$$${but}\:{grad}\:{of}\:{l}_{\mathrm{1}} ×{l}_{\mathrm{2}} =−\mathrm{1} \\ $$$${grad}\:{of}\:{l}_{\mathrm{2}} =−\mathrm{1} \\ $$$${equation}\:{of}\:{the}\:{perpendicular}\:{line}\:{l}_{\mathrm{2}} \\ $$$$\frac{{y}−\mathrm{5}}{{x}−\mathrm{2}}=−\mathrm{1} \\ $$$${y}−\mathrm{5}=−{x}+\mathrm{2} \\ $$$${y}=−{x}+\mathrm{7} \\ $$