Question

Prove that $(-5,-3),(1,-11),(7,-6),(1,2)$ coordinates are the vertices of parallelograms

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 $A=(-5,-3)$, $B=(1,-11)$, $C=(7,-6)$ and $D=(1,2)$

We first find the distance between $A=(-5,-3)$ and $B=(1,-11)$ as follows:

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(1-(-5\left)\right)}^2+{(-11-(-3\left)\right)}^2}=\sqrt{{(1+5)}^2+{(-11+3)}^2}=\sqrt{6^2+{(-8)}^2}$

$=\sqrt{36+64}=\sqrt{100}=10$

Similarly, the distance between $B=(1,-11)$ and $C=(7,-6)$ is:

$BC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(7-1)}^2+{(-6-(-11\left)\right)}^2}=\sqrt{6^2+{(-6+11)}^2}=\sqrt{6^2+5^2}=\sqrt{36+25}$

$=\sqrt{61}$

Now, the distance between $C=(7,-6)$ and $D=(1,2)$ is:

$CD=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(1-7)}^2+{(2-(-6\left)\right)}^2}=\sqrt{(-6{)}^2+{(2+6)}^2}=\sqrt{(-6{)}^2+8^2}=\sqrt{36+64}$

$=\sqrt{100}=10$

Now, the distance between $D=(1,2)$ and $A=(-5,-3)$ is:

$DA=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}=\sqrt{{(1-(-5\left)\right)}^2+{(2-(-3\left)\right)}^2}=\sqrt{{(1+5)}^2+{(2+3)}^2}=\sqrt{6^2+5^2}=\sqrt{36+25}$

$=\sqrt{61}$

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:

$AB=CD=10$ and $BC=DA=\sqrt{61}$

Hence, the given vertices are the vertices of parallelogram.