A triangle with vertices (4, 0), (-1, -1), (3, 5) is
- isosceles and right angled
- isosceles but not right angled
- right angled but not isosceles
- neither right angled nor isoceles
Solution
$A B=\sqrt{(4+1)^2+(0+1)^2}=\sqrt{26} ; B C=\sqrt{(3+1)^2+(5+1)^2}=\sqrt{52}$ $C A=\sqrt{(4-3)^2+(0-5)^2}=\sqrt{26} ;$ 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
Asked in: JEE Main 2002