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Prove that the points $(4, 5), (7, 6), (6, 3), (3, 2)$ are the vertices of a parallelogram. Is it a rectangle?
Given:
Given vertices are $(4, 5), (7, 6), (6, 3), (3, 2)$.
To do:
We have to prove that the points $(4, 5), (7, 6), (6, 3), (3, 2)$ are the vertices of a parallelogram and check whether it is a rectangle.
Solution:
Let the vertices of the parallelogram be $A(4, 5), B(7, 6), C(6, 3)$ and $D(3, 2)$.
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,
\( \mathrm{AB}=\sqrt{(7-4)^{2}+(6-5)^{2}} \)
\( =\sqrt{(3)^{2}+(1)^{2}} \)
\( =\sqrt{9+1} \)
\( =\sqrt{10} \)
\( \mathrm{BC}=\sqrt{(6-7)^{2}+(3-6)^{2}} \)
\( =\sqrt{(-1)^{2}+(-3)^{2}} \)
\( =\sqrt{1+9} \)
\( =\sqrt{10} \)
\( \mathrm{CD}=\sqrt{(3-6)^{2}+(2-3)^{2}} \)
\( =\sqrt{(-3)^{2}+(-1)^{2}} \)
\( =\sqrt{9+1} \)
\( =\sqrt{10} \)
\( \mathrm{DA}=\sqrt{(4-3)^{2}+(5-2)^{2}} \)
\( =\sqrt{(1)^{2}+(3)^{2}} \)
\( =\sqrt{1+9} \)
\( =\sqrt{10} \)
\( \mathrm{AC}=\sqrt{(6-4)^{2}+(3-5)^{2}} \)
\( =\sqrt{(2)^{2}+(-2)^{2}} \)
\( =\sqrt{4+4} \)
\( =\sqrt{8} \)
\( \mathrm{BD}=\sqrt{(3-7)^{2}+(2-6)^{2}} \)
\( =\sqrt{(-4)^{2}+(-4)^{2}} \)
\( =\sqrt{16+16} \)
\( =\sqrt{32} \)
Here,
\( \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \)
\( \mathrm{AC}≠\mathrm{BD} \)
Therefore, $(4, 5), (7, 6), (6, 3)$ and $(3, 2)$ are the vertices of a parallelogram. It is not a rectangle.
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