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Show that the points $A (1, -2), B (3, 6), C (5, 10)$ and $D (3, 2)$ are the vertices of a parallelogram.
Given:
Given vertices are $A (1, -2), B (3, 6), C (5, 10)$ and $D (3, 2)$.
To do:
We have to show that the points $A (1, -2), B (3, 6), C (5, 10)$ and $D (3, 2)$ are the vertices of a parallelogram.
Solution:
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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(3-1)^{2}+(6+2)^{2}}=\sqrt{(2)^{2}+(8)^{2}} \)
\( =\sqrt{4+64}=\sqrt{68} \)
Similarly,
\( \mathrm{BC}=\sqrt{(5-3)^{2}+(10-6)^{2}} \)
\( =\sqrt{(2)^{2}+(4)^{2}}=\sqrt{4+16}=\sqrt{20} \)
\( \mathrm{CD}=\sqrt{(3-5)^{2}+(2-10)^{2}} \)
\( =\sqrt{(-2)^{2}+(-8)^{2}}=\sqrt{4+64}=\sqrt{68} \)
\( \mathrm{DA}=\sqrt{(3-1)^{2}+(2+2)^{2}} \)
\( =\sqrt{(2)^{2}+(4)^{2}}=\sqrt{4+16}=\sqrt{20} \)
Here, \( \mathrm{AB}=\mathrm{CD} \) and \( \mathrm{AD}=\mathrm{BC} \)
Therefore, $ABCD$ is a parallelogram.
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