Question

Show that A(−3, −1), B(−4, −1), C(3, 3) and D(4, 3) are the vertices of a rhombus.

Answer 1

Disclaimer:- It should be parallelogram in place of rhombus in the question.
The given vertices are A(−3, −1), B(−4, −1), C(3, 3) and D​(4, 3).
Then,
$$\begin{gathered} AB=\sqrt{{\left(-4+3\right)}^2+{\left(-1+1\right)}^2} \\ =\sqrt{{\left(-1\right)}^2+{\left(0\right)}^2} \\ =\sqrt{1+0} \\ =1\text{unit} \\ BC=\sqrt{{\left(3+4\right)}^2+{\left(3+1\right)}^2} \\ =\sqrt{{\left(7\right)}^2+{\left(4\right)}^2} \\ =\sqrt{49+16} \\ =\sqrt{65}\text{units} \\ CD=\sqrt{{\left(4-3\right)}^2+{\left(3-3\right)}^2} \\ =\sqrt{{\left(1\right)}^2+{\left(0\right)}^2} \\ =\sqrt{1+0} \\ =1\text{unit} \\ DA=\sqrt{{\left(-3-4\right)}^2+{\left(-1-3\right)}^2} \\ =\sqrt{{\left(-7\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{49+16} \\ =\sqrt{65}\text{units} \end{gathered}$$
Here, AB = CD
and BC = DA
Therefore, ABCD is a parallelogram.