Question Number 39127 by Rio Mike last updated on 02/Jul/18
$${Given}\:{the}\:{lines} \\ $$$${l}_{\mathrm{1}} :\:\:{x}\:+\:{y}\:=\:\mathrm{5}\:{and}\:{l}_{\mathrm{2}} :\:{y}\:=\:\mathrm{4}{x} \\ $$$${and}\:{l}_{\mathrm{3}} ;\:\mathrm{4}{x}\:+\:{y}\:−\:\mathrm{1}\:=\mathrm{0} \\ $$$${show}\:{that}\:{l}_{\mathrm{2}\:} \:{is}\:{perpendicular} \\ $$$${to}\:{l}_{\mathrm{3}} . \\ $$$${find}\:{the}\:{point}\:{coordinates} \\ $$$${if}\:{x}\:+\:\mathrm{2}{y}\:=\:\mathrm{5}\:{is}\:{colliner}\:{to}\:{l}_{\mathrm{1}} \\ $$
Commented by mondodotto@gmail.com last updated on 03/Jul/18
$$\boldsymbol{{two}}\:\boldsymbol{{lines}}\:\boldsymbol{{are}}\:\boldsymbol{{said}}\:\boldsymbol{{to}}\:\boldsymbol{{be}}\:\boldsymbol{{perpindicular}}\:\boldsymbol{{if}}\:\boldsymbol{{product}}\:\boldsymbol{{of}}\:\boldsymbol{{their}}\: \\ $$$$\boldsymbol{{gradient}}\:\boldsymbol{{equals}}\:\left(−\mathrm{1}\right) \\ $$$$\boldsymbol{{from}}\:\boldsymbol{{l}}_{\mathrm{2}} \:\boldsymbol{{gradient}}\:\boldsymbol{{is}}\:\mathrm{4}\:\boldsymbol{{and}} \\ $$$$\boldsymbol{{l}}_{\mathrm{3}} \:\boldsymbol{{gradient}}\:\boldsymbol{{is}}\:\left(−\mathrm{4}\right)\:\boldsymbol{{then}} \\ $$$$\boldsymbol{{product}}\:\boldsymbol{{of}}\:\boldsymbol{{these}}\:\boldsymbol{{gradient}}\:\boldsymbol{{is}}\:\boldsymbol{{not}}\:\left(−\mathrm{1}\right) \\ $$$$\therefore\:\boldsymbol{{l}}_{\mathrm{2}\:} \:\boldsymbol{{and}}\:\boldsymbol{{l}}_{\mathrm{3}} \:\boldsymbol{{not}}\:\boldsymbol{{perpendicular}}\:\boldsymbol{{lines}} \\ $$