• # question_answer Prove that the points (3, 0), (6, 4) and ($1,\text{ }3$) are the vertices of a right angled isosceles triangle.

 Let $A(3,0),\text{ }B(6,4)$ and $C(-1,3)$ be the vertices of a triangle $ABC$. Length of                       $AB=\sqrt{{{(6-3)}^{2}}+{{(4-0)}^{2}}}$ $=\sqrt{{{(3)}^{2}}+{{(4)}^{2}}}$ $=\sqrt{9+16}=\sqrt{25}=5\,$units Length of                       $BC=\sqrt{{{(-1-6)}^{2}}+{{(3-4)}^{2}}}$ $=\sqrt{{{(-7)}^{2}}+{{(-1)}^{2}}}$ $=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$ units. And Length of    $AC=\sqrt{{{(-1-3)}^{2}}+{{(3-0)}^{2}}}$ $=\sqrt{{{(-4)}^{2}}+{{(3)}^{2}}}$ $=\sqrt{16+9}=\sqrt{25}=5$ units $\therefore$                         $AB=AC$ And         ${{(AB)}^{2}}+{{(AC)}^{2}}={{(BC)}^{2}}$ Hence, $\Delta \text{ }ABC$ is a isosceles, right angled triangle.                              Hence Proved