# Prove that the points $(3, -2), (4, 0), (6, -3)$ and $(5, -5)$ are the vertices of a parallelogram.

Given:

Given vertices are $(3, -2), (4, 0), (6, -3)$ and $(5, -5)$.

To do:

We have to show that the points $(3, -2), (4, 0), (6, -3)$ and $(5, -5)$ are the vertices of a parallelogram.

Solution:

Let the vertices of the parallelogram be $A(3, -2), B(4, 0), C(6, -3)$ and $D(5, -5)$.

We know that,

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

$\mathrm{AB}=\sqrt{(4-3)^{2}+(0+2)^{2}}$

$=\sqrt{(1)^{2}+(2)^{2}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

$\mathrm{BC}=\sqrt{(6-4)^{2}+(-3-0)^{2}}$

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

$=\sqrt{4+9}$

$=\sqrt{13}$

$\mathrm{CD}=\sqrt{(5-6)^{2}+(-5+3)^{2}}$

$=\sqrt{(-1)^{2}+(-2)^{2}}$

$=\sqrt{1+4}$

$=\sqrt{5}$

$\mathrm{DA}=\sqrt{(5-3)^{2}+(-5+2)^{2}}$

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

$=\sqrt{4+9}$

$=\sqrt{13}$

Here, $\mathrm{AB}=\mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}$

Therefore, $(3, -2), (4, 0), (6, -3)$ and $(5, -5)$ are the vertices of a parallelogram.

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Updated on: 10-Oct-2022

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