Question

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

Answer 1

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).
$$\begin{gathered} AB=\sqrt{{\left(1-4\right)}^2+{\left(2-3\right)}^2}=\sqrt{{\left(-3\right)}^2+{\left(-1\right)}^2} \\ =\sqrt{9+1}=\sqrt{10} \end{gathered}$$
$$\begin{gathered} BC=\sqrt{{\left(4-6\right)}^2+{\left(3-6\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(-3\right)}^2} \\ =\sqrt{4+9}=\sqrt{13} \end{gathered}$$
$$\begin{gathered} CD=\sqrt{{\left(6-3\right)}^2+{\left(6-5\right)}^2}=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2} \\ =\sqrt{9+1}=\sqrt{10} \\ AD=\sqrt{{\left(1-3\right)}^2+{\left(2-5\right)}^2}=\sqrt{{\left(-2\right)}^2+{\left(-3\right)}^2} \\ =\sqrt{4+9}=\sqrt{13} \end{gathered}$$
$\because AB=CD=\sqrt{10}\mathrm{units}\mathrm{and}BC=AD=\sqrt{13}\mathrm{units}$
Therefore, ABCD is a parallelogram. Now
$$\begin{gathered} AC=\sqrt{{\left(1-6\right)}^2+{\left(2-6\right)}^2}=\sqrt{{\left(-5\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{25+16}=\sqrt{41} \\ BD=\sqrt{{\left(4-3\right)}^2+{\left(3-5\right)}^2}=\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2} \\ =\sqrt{1+4}=\sqrt{5} \end{gathered}$$
Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.