Question Number 191169 by Mingma last updated on 19/Apr/23
Answered by mehdee42 last updated on 19/Apr/23
$${c}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{m}_{{l}} ={y}'_{{A}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{l}:\:\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\mathrm{4}\right)\: \\ $$$${m}_{{d}} ×{m}_{{l}} =−\mathrm{1}\Rightarrow{m}_{{l}} =−\sqrt{\mathrm{3}}\Rightarrow{d}:\:\:\:{y}=−\sqrt{\mathrm{3}}\left({x}−\mathrm{2}\right) \\ $$$$\left({l}\right)\:,\:\left({d}\:\right)\:\Rightarrow\:{H}\left(\frac{\mathrm{5}}{\mathrm{2}},−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$
Commented by mehdee42 last updated on 19/Apr/23
Commented by Mingma last updated on 19/Apr/23
Nice solution, sir!