• # 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

(i) A triangle is isosceles, if its any two sides are equal, so to prove a triangle isosceles, we have to prove as two sides are equal. (ii) To prove a triangle a right angled triangle, we have to prove that the sum of square of any two sides is equal to the square of third side. Let points A (4, 0), B (-1, - 1) and C(3, 5) be the vertices of a $\Delta ABC$. $\therefore AB=\sqrt{(-1-{{4}^{2}}+{{(-1-0)}^{2}}}=\sqrt{25+1}=\sqrt{26}$    $BC=\sqrt{{{(3+1)}^{2}}+{{(5+1)}^{2}}}$    $=\sqrt{{{4}^{2}}+{{6}^{2}}}$    $=\sqrt{16+36}=\sqrt{52}$ and $CA=\sqrt{{{(4-3)}^{2}}+{{(0-5)}^{2}}}=\sqrt{1+25}=\sqrt{26}$ Now, $C{{A}^{2}}+A{{B}^{2}}={{(\sqrt{26})}^{2}}+{{(\sqrt{26})}^{2}}$                                    $=26+26=52=B{{C}^{2}}$ $\Rightarrow$   $C{{A}^{2}}+A{{B}^{2}}=B{{C}^{2}}$ Thus, the triangle is isosceles and right angled triangle.