Question

Show that the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of a right-angled triangle. Also, prove that these are the vertices of an isosceles triangle.

Answer 1

A(3, 0), B (6, 4) and C(−1, 3) are the given points. Then:
$$\begin{gathered} AB=\sqrt{{\left(6-3\right)}^2+{\left(4-0\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5\mathrm{units} \\ BC=\sqrt{{\left(-1-6\right)}^2+{\left(3-4\right)}^2}=\sqrt{{\left(-7\right)}^2+{\left(-1\right)}^2}=\sqrt{49+1}=\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\mathrm{units} \\ AC=\sqrt{{\left(-1-3\right)}^2+{\left(3-0\right)}^2}=\sqrt{{\left(-4\right)}^2+{\left(3\right)}^2}=\sqrt{16+9}=\sqrt{25}=5\mathrm{units} \end{gathered}$$
Thus, AB = BC = 5 units
Also, (AB)²+(AC)² = (5)²+(5)² = 50
and (BC)² = (5$\sqrt{2}$)² = 25$\times$2 = 50
Thus, (AB)²+(AC)² = (BC)²
This show that $∆ABC$ is right- angled at A.
This proves that the the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of an isosceles right- angled triangle.