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A triangle with vertices (4, 0), (-1, -1), (3, 5) is
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isosceles and right angled
isosceles and right angled
$A B=\sqrt{(4+1)^2+(0+1)^2}=\sqrt{26} ; B C=\sqrt{(3+1)^2+(5+1)^2}=\sqrt{52}$
$C A=\sqrt{(4-3)^2+(0-5)^2}=\sqrt{26} ;$ So, in isosceles triangle $A B=C A$
For right angled triangle $\mathrm{BC}^2=\mathrm{AB}^2+\mathrm{AC}^2$
So, here $B C=\sqrt{52}$ or $B C^2=52$ or $(\sqrt{26})^2+(\sqrt{26})^2=52$
So, given triangle is right angled and also isosceles
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