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Prove that the points $A (2, 3), B (-2, 2), C (-1, -2)$ and $D (3, -1)$ are the vertices of a square $ABCD$.
Given:
Given points are $A (2, 3), B (-2, 2), C (-1, -2)$ and $D (3, -1)$.
To do:
We have to prove that the points $A (2, 3), B (-2, 2), C (-1, -2)$ and $D (3, -1)$ are the vertices of a square.
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,
\( A B=\sqrt{(-2-2)^{2}+(2-3)^{2}} \)
\( =\sqrt{(-4)^{2}+(-1)^{2}} \)
\( =\sqrt{16+1} \)
\( =\sqrt{17} \)
\( B C=\sqrt{(-1+2)^{2}+(-2-2)^{2}} \)
\( =\sqrt{(1)^{2}+(-4)^{2}} \)
\( =\sqrt{1+16} \)
\( =\sqrt{17} \)
\( C D=\sqrt{(3+1)^{2}+(-1+2)^{2}} \)
\( =\sqrt{16+1} \)
\( =\sqrt{17} \)
\( D A=\sqrt{(2-3)^{2}+(3+1)^{2}} \)
\( =\sqrt{(-1)^{2}+(4)^{2}} \)
\( =\sqrt{1+16} \)
\( =\sqrt{17} \)
\( A C=\sqrt{(-1-2)^{2}+(-2-3)^{2}} \)
\( =\sqrt{(-3)^{2}+(-5)^{2}} \)
\( =\sqrt{9+25} \)
\( =\sqrt{34} \)
\( B D=\sqrt{(3+2)^{2}+(-1-2)^{2}} \)
\( =\sqrt{(5)^{2}+(-3)^{2}} \)
\( =\sqrt{25+9} \)
\( =\sqrt{34} \)
\( \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \) 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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