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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