Question Number 210842 by peter frank last updated on 20/Aug/24
Answered by mm1342 last updated on 20/Aug/24
$${AB}:{y}−\mathrm{2}{x}=\mathrm{1} \\ $$$${AC}:{y}−{x}=\mathrm{0} \\ $$$${BC}:−\mathrm{3}{y}+{x}=−\mathrm{4} \\ $$$$\Rightarrow{A}\left(−\mathrm{1},−\mathrm{1}\right)\:\&\:{B}\left(\frac{\mathrm{1}}{\mathrm{5}},\frac{\mathrm{7}}{\mathrm{5}}\right)\:\&\:{C}\left(\mathrm{2},\mathrm{2}\right)\Rightarrow{D}\left(\frac{\mathrm{11}}{\mathrm{10}},\frac{\mathrm{17}}{\mathrm{10}}\right) \\ $$$$\Rightarrow{AB}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{5}}\:\:\&\:{AC}^{\mathrm{2}} =\mathrm{18}\:\:\&\:\:{AD}^{\mathrm{2}} =\frac{\mathrm{117}}{\mathrm{10}}\:\&\:{BD}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} =\frac{\mathrm{126}}{\mathrm{5}}\:\:\&\:{AD}^{\mathrm{2}} +{BD}^{\mathrm{2}} =\frac{\mathrm{126}}{\mathrm{10}} \\ $$$$\Rightarrow\checkmark \\ $$$$ \\ $$$$ \\ $$