Question Number 77514 by liki last updated on 07/Jan/20
Answered by jagoll last updated on 07/Jan/20
$${L}_{\mathrm{1}} :\:{y}=−\frac{\mathrm{3}}{\mathrm{4}}\left({x}−\mathrm{2}\right)+\left(−\mathrm{1}\right) \\ $$$${L}_{\mathrm{1}} :\:{y}\:=\:−\frac{\mathrm{3}}{\mathrm{4}}{x}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{a}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${let}\:{poin}\:{C}\left({x},{y}\right)\:,\:{CA}\:{perpendicular} \\ $$$${to}\:{L}_{\mathrm{1}} \:\Rightarrow\:{y}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{4}}{\mathrm{3}}{x}\: \\ $$$${y}\:=\:\frac{\mathrm{4}}{\mathrm{3}}{x}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by liki last updated on 07/Jan/20
$$.\boldsymbol{{Thanks}}\:\boldsymbol{{alot}}\:\boldsymbol{{sir}}. \\ $$
Answered by Kunal12588 last updated on 07/Jan/20
$${let}\:{coordinates}\:{of}\:{point}\:{C}\:{be}\:\left({h},{k}\right) \\ $$$${L}_{\mathrm{1}} :\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{2}\: \\ $$$$\because\:{A}\left(\mathrm{0},{a}\right)\:{satisfy}\:{L}_{\mathrm{1}} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${CA}\::\:\mathrm{6}{y}−\mathrm{8}{x}=\mathrm{3} \\ $$$${L}_{\mathrm{2}} \::\:{x}=\mathrm{2} \\ $$$$\because\:{C}\left({h},{k}\right)\:{satisfy}\:{CA}\:{and}\:{L}_{\mathrm{2}} \\ $$$$\therefore\:\mathrm{6}{k}−\mathrm{8}{h}=\mathrm{3}\:{and}\:{h}=\mathrm{2} \\ $$$$\Rightarrow{k}=\frac{\mathrm{19}}{\mathrm{6}} \\ $$$${point}\:{C}\left(\mathrm{2}\:,\:\frac{\mathrm{19}}{\mathrm{6}}\right) \\ $$
Commented by liki last updated on 07/Jan/20
$$.\boldsymbol{{T}}{hanks}\:{so}\:{much}\:{sir}. \\ $$