Question

Show that the following points taken in order form the vertices of a parallelogram

(-7, -5), (-4, 3), (5, 6) and (2, -2)

Answer 1

Let $A,B,C$ and $D$ represent the points $(–7,–5),(–4,3),(5,6)$ and $(2,–2)$

Distance of $AB$

$AB=\sqrt{(–4+7{)}^2+(3+5{)}^2}$

$AB=\sqrt{3^2+8^2}$

$AB=\sqrt{9+64}$

$AB=\sqrt{73}$

Similarly

Distance of $BC$

$BC=\sqrt{(5+4{)}^2+(6-3{)}^2}$

$BC=\sqrt{9^2+3^2}$

$BC=\sqrt{81+9}$

$BC=\sqrt{90}$

Similarly

Distance of $CD$

$CD=\sqrt{(2-5{)}^2+(-2-6{)}^2}$

$CD=\sqrt{(-3{)}^2+(-8{)}^2}$

$CD=\sqrt{9+64}$

$CD=\sqrt{73}$

Similarly

Distance of $DA$

$DA=\sqrt{(-7-2{)}^2+(-5+2{)}^2}$

$DA=\sqrt{(-9{)}^2+(-3{)}^2}$

$DA=\sqrt{81+9}$

$DA=\sqrt{90}$

So, $AB=CD=\sqrt{73}$ and $BC=DA=\sqrt{90}$

i.e., The opposite sides are equal.

Hence $ABCD$ is a parallelogram.