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Question-210842




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

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