Question

The co-ordinates of □EFGH are E(3, −2), F(7, −2), G(5, −5) and H(1, −5) respectively. Draw □EFGH and state its type.

Answer 1

We know that distance between the two points (x₁ ,y₁ ) and (x₂ ,y₂ ) is $\sqrt{{\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2}$.
EF=$\sqrt{{\left(-2+2\right)}^2+{\left(7-3\right)}^2}=4$ units
GH= $\sqrt{{\left(-5+5\right)}^2+{\left(5-1\right)}^2}=4$ units
EH= $\sqrt{{\left(-5+2\right)}^2+{\left(1-3\right)}^2}=\sqrt{13}$ units
FG=$\sqrt{{\left(-5+2\right)}^2+{\left(5-7\right)}^2}=\sqrt{13}$ units
As EF = GH and EH = FG, the opposite sides are equal.
Therefore, the quadrilateral EFGH is a parallelogram.