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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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