Question

Show that points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are vertices of a rhombus ABCD.

Answer 1

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
$$\begin{gathered} \mathrm{AB}=\sqrt{{\left(-4-\left(-1\right)\right)}^2+{\left(-7-2\right)}^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10} \\ \mathrm{BC}=\sqrt{{\left(-1-8\right)}^2+{\left(2-5\right)}^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10} \\ \mathrm{CD}=\sqrt{{\left(8-5\right)}^2+{\left(5-\left(-4\right)\right)}^2}=\sqrt{9+81}=\sqrt{90}=3\sqrt{10} \\ \mathrm{DA}=\sqrt{{\left(5-\left(-4\right)\right)}^2+{\left(-4-\left(-7\right)\right)}^2}=\sqrt{81+9}=\sqrt{90}=3\sqrt{10} \end{gathered}$$
Thus, all the sides are equal.
$$\begin{gathered} \mathrm{AC}=\sqrt{{\left(-4-8\right)}^2+{\left(-7-5\right)}^2}=\sqrt{144+144}=12\sqrt{2} \\ \mathrm{DB}=\sqrt{{\left(8-5\right)}^2+{\left(2-\left(-4\right)\right)}^2}=\sqrt{9+36}=3\sqrt{5} \end{gathered}$$
Thus, the diagonals are not equal.
So, the given vertices form a rhombus ABCD.