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Show that the points $ (2,2),(-2,2) $, $ (-2,-2) $ and $ (2,-2) $ are the vertices of a square.
Given:
Given points are \( (2,2),(-2,2) \), \( (-2,-2) \) and \( (2,-2) \)
To do:
We have to show that the points \( (2,2),(-2,2) \), \( (-2,-2) \) and \( (2,-2) \) are the vertices of a square.
Solution:
Let $A=(2,2), B=(-2,2), C=(-2,-2)$ and $D=(2,-2)$
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,
\( A B=\sqrt{(-2-2)^{2}+(2-2)^{2}} \)
\( =\sqrt{(-4)^{2}+(0)^{2}} \)
\( =\sqrt{16+0} \)
\( =4 \)
\( B C=\sqrt{(-2+2)^{2}+(-2-2)^{2}} \)
\( =\sqrt{(0)^{2}+(-4)^{2}} \)
\( =\sqrt{0+16} \)
\( =4 \)
\( C D=\sqrt{(2+2)^{2}+(-2+2)^{2}} \)
\( =\sqrt{16+0} \)
\( =4 \)
\( D A=\sqrt{(2-2)^{2}+(2+2)^{2}} \)
\( =\sqrt{(0)^{2}+(4)^{2}} \)
\( =\sqrt{16} \)
\( =4 \)
\( A C=\sqrt{(-2-2)^{2}+(-2-2)^{2}} \)
\( =\sqrt{(-4)^{2}+(-4)^{2}} \)
\( =\sqrt{16+16} \)
\( =\sqrt{32} \)
\( B D=\sqrt{(2+2)^{2}+(-2-2)^{2}} \)
\( =\sqrt{(4)^{2}+(-4)^{2}} \)
\( =\sqrt{16+16} \)
\( =\sqrt{32} \)
\( \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \) and \( \mathrm{AC}=\mathrm{BD} \)
Here, all the sides are equal and the diagonals are equal to each other.
Therefore, the given points are the vertices of a square.
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