question_answer

A triangle with vertices\[(4,\,\,0),\,\,(-1,\,\,-1),\,\,(3,\,\,5)\] is:

A)  isosceles and right angled

B)  isosceles but not right angled

C)  right angled but not isosceles

D)  neither right angled nor isosceles

Correct Answer: A

Solution :

\[AB=\sqrt{{{(4+1)}^{2}}+{{(0+1)}^{2}}}=\sqrt{26}\] \[BC=\sqrt{{{(3+1)}^{2}}+{{(5+1)}^{2}}}=\sqrt{52}\] \[CA=\sqrt{{{(4-3)}^{2}}+{{(0-5)}^{2}}}=\sqrt{26}\] So, in isosceles triangles side\[AB=CA\] For right angled triangle                 \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\] So, here \[BC=\sqrt{52}=B{{C}^{2}}=52\] or     \[{{(\sqrt{26})}^{2}}={{(\sqrt{26})}^{2}}=52\] So, given vertices is right angled and also is isosceles triangle.