Question

Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.

Answer 1

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).
$$\begin{gathered} AB=\sqrt{{\left(-5+3\right)}^2+{\left(-5-2\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(-7\right)}^2}=\sqrt{4+49}=\sqrt{53}\mathrm{units}. \\ BC=\sqrt{{\left(2+5\right)}^2+{\left(-3+5\right)}^2}=\sqrt{{\left(7\right)}^2+{\left(2\right)}^2}=\sqrt{49+4}=\sqrt{53}\mathrm{units}. \\ CD=\sqrt{{\left(4-2\right)}^2+{\left(4+3\right)}^2}=\sqrt{{\left(2\right)}^2+{\left(7\right)}^2}=\sqrt{4+49}=\sqrt{53}\mathrm{units}. \\ DA\hspace{0.17em}=\sqrt{{\left(4+3\right)}^2+{\left(4-2\right)}^2}=\sqrt{{\left(7\right)}^2+{\left(2\right)}^2}=\sqrt{49+4}=\sqrt{53}\mathrm{units}. \\ \mathrm{Therefore},AB\hspace{0.17em}=BC\hspace{0.17em}=CD=DA=\sqrt{53}\mathrm{units} \\ AC=\sqrt{{\left(2+3\right)}^2+{\left(-3-2\right)}^2}=\sqrt{{\left(5\right)}^2+{\left(-5\right)}^2}=\sqrt{25+25}=\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}\mathrm{units} \\ BD=\sqrt{{\left(4+5\right)}^2+{\left(4+5\right)}^2}=\sqrt{{\left(9\right)}^2+{\left(9\right)}^2}=\sqrt{81+81}=\sqrt{162}=\sqrt{81\times 2}=9\sqrt{2}\mathrm{units} \\ \mathrm{Thus},\mathrm{diagonal}AC\mathrm{is}\mathrm{not}\mathrm{equal}\mathrm{to}\mathrm{diagonal}BD. \end{gathered}$$
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.
$$\begin{gathered} \mathrm{Area}\mathrm{of}\text{a}\mathrm{rhombus}=\frac{1}{2}\times \left(\mathrm{product}\mathrm{of}\mathrm{diagonals}\right) \\ =\frac{1}{2}\times \left(5\sqrt{2}\right)\times \left(9\sqrt{2}\right) \\ =\frac{45\left(2\right)}{2} \\ =45\mathrm{square}\mathrm{units} \end{gathered}$$