Mathematics - Point Question with Solution | TestHub

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