Question

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)

Answer 1

Let the points (- 1, - 2), (1, 0), ( - 1, 2) and ( - 3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.
Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore AB=\sqrt{{(-1-1)}^2+{(-2-0)}^2}$
$=\sqrt{{(-2)}^2+{(-2)}^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
$BC=\sqrt{{(1-(-1))}^2+{(0-2)}^2}$
$=\sqrt{{\left(2\right)}^2+{(-2)}^2}=\sqrt{4+4}$
$=\sqrt{8}=2\sqrt{2}$
$CD=\sqrt{{(-1-(-3))}^2+{(2-0)}^2}$
$=\sqrt{{\left(2\right)}^2+{\left(2\right)}^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
$AD=\sqrt{{(-1-(-3))}^2+{(-2-0)}^2}$
$=\sqrt{{\left(2\right)}^2+{(-2)}^2}$
$=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
Diagonal $AC=\sqrt{{(-1-(-1))}^2+{(-2-2)}^2}$
$=\sqrt{0^2+{(-4)}^2}=\sqrt{16}=4$
Diagonal $BD=\sqrt{{(1-(-3))}^2+{(0-0)}^2}$
$=\sqrt{4^2+0^2}=\sqrt{16}=4$
It can be observed that all sides of this quadrilateral are of same length and also, the diagonals are of same length. Therefore, the given points are the vertices of a square.