Question

Prove that the set of coordinates are the vertices of parallelogram $(4,0),(-2,-3),(3,2),(-3,-1)$.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Let the given vertices be $A=(4,0)$, $B=(-2,-3)$, $C=(3,2)$ and $D=(-3,-1)$

We first find the distance between $A=(4,0)$ and $B=(-2,-3)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-2-4)}^2+{(-3-0)}^2}=\sqrt{{(-6)}^2+{(-3)}^2}=\sqrt{36+9}=\sqrt{45}=\sqrt{3^2\times 5}$

$=3\sqrt{5}$

Similarly, the distance between $B=(-2,-3)$ and $C=(3,2)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(3-(-2\left)\right)}^2+{(2-(-3\left)\right)}^2}=\sqrt{{(3+2)}^2+{(2+3)}^2}=\sqrt{5^2+5^2}=\sqrt{25+25}$
$=\sqrt{50}=\sqrt{5^2\times 2}=5\sqrt{2}$

Now, the distance between $C=(3,2)$ and $D=(-3,-1)$ is:

$CD=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-3-3)}^2+{(-1-2)}^2}=\sqrt{{(-6)}^2+{(-3)}^2}=\sqrt{36+9}=\sqrt{45}=\sqrt{3^2\times 5}$

$=3\sqrt{5}$

Now, the distance between $D=(-3,-1)$ and $A=(4,0)$ is:

$DA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(-3-4)}^2+{(-1-0)}^2}=\sqrt{{(-7)}^2+{(-1)}^2}=\sqrt{49+1}=\sqrt{50}=\sqrt{5^2\times 2}$
$=5\sqrt{2}$

We also know that if the opposite sides have equal side lengths, then $ABCD$ is a parallelogram.

Here, since the lengths of the opposite sides are equal that is:

$AB=CD=3\sqrt{5}$ and $BC=DA=5\sqrt{2}$

Hence, the given vertices are the vertices of parallelogram.