Question

Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

Answer 1

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).
$$\begin{gathered} AB=\sqrt{{\left(3-\left(-5\right)\right)}^2+{\left(0-6\right)}^2}=\sqrt{{\left(8\right)}^2+{\left(-6\right)}^2}=\sqrt{64+36}=\sqrt{100}=10\mathrm{units} \\ BC=\sqrt{{\left(9-3\right)}^2+{\left(8-0\right)}^2}=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=\sqrt{36+64}=\sqrt{100}=10\mathrm{units} \\ AC=\sqrt{{\left(9-\left(-5\right)\right)}^2+{\left(8-6\right)}^2}=\sqrt{{\left(14\right)}^2+{\left(2\right)}^2}=\sqrt{196+4}=\sqrt{200}=10\sqrt{2}\mathrm{units} \\ \mathrm{Therefore},AB\hspace{0.17em}=BC=10\mathrm{units} \end{gathered}$$

Also, (AB)²+(BC)² = ${\left(10\right)}^2+{\left(10\right)}^2=200$
and (AC)² = ${\left(10\sqrt{2}\right)}^2=200$
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ABC$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, $\text{a}\mathrm{rea}\text{of a triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$
$$\begin{gathered} \mathrm{If}AB\mathrm{is}\mathrm{the}\mathrm{height}\mathrm{and}BC\mathrm{is}\mathrm{the}\mathrm{base}, \\ \mathrm{Area}=\frac{1}{2}\times 10\times 10 \\ =50\mathrm{square}\mathrm{units} \end{gathered}$$