Question

Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle.

Answer 1

The given points are A(7, 10), B(−2, 5) and C(3, −4).
$$\begin{gathered} AB=\sqrt{{\left(-2-7\right)}^2+{\left(5-10\right)}^2}=\sqrt{{\left(-9\right)}^2+{\left(-5\right)}^2}=\sqrt{81+25}=\sqrt{106} \\ BC=\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(-4-5\right)}^2}=\sqrt{{\left(5\right)}^2+{\left(-9\right)}^2}=\sqrt{25+81}=\sqrt{106} \\ AC=\sqrt{{\left(3-7\right)}^2+{\left(-4-10\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(-14\right)}^2}=\sqrt{16+196}=\sqrt{212} \end{gathered}$$
Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)²+(BC)² = ${\left(\sqrt{106}\right)}^2+{\left(\sqrt{106}\right)}^2=212$
and (AC)² = ${\left(\sqrt{212}\right)}^2$ = 212
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.