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Show that the points $A (5, 6), B (1, 5), C (2, 1)$ and $D (6, 2)$ are the vertices of a square.
Given:
Given points are $A (5, 6), B (1, 5), C (2, 1)$ and $D (6, 2)$.
To do:
We have to show that the points $A (5, 6), B (1, 5), C (2, 1)$ and $D (6, 2)$ 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,
\( A B=\sqrt{(1-5)^{2}+(5-6)^{2}} \)
\( =\sqrt{(-4)^{2}+(-1)^{2}} \)
\( =\sqrt{16+1} \)
\( =\sqrt{17} \)
\( B C=\sqrt{(2-1)^{2}+(1-5)^{2}} \)
\( =\sqrt{(1)^{2}+(-4)^{2}} \)
\( =\sqrt{1+16} \)
\( =\sqrt{17} \)
\( C D=\sqrt{(6-2)^{2}+(2-1)^{2}} \)
\( =\sqrt{16+1} \)
\( =\sqrt{17} \)
\( D A=\sqrt{(5-6)^{2}+(6-2)^{2}} \)
\( =\sqrt{(-1)^{2}+(4)^{2}} \)
\( =\sqrt{1+16} \)
\( =\sqrt{17} \)
\( A C=\sqrt{(2-5)^{2}+(1-6)^{2}} \)
\( =\sqrt{(-3)^{2}+(-5)^{2}} \)
\( =\sqrt{9+25} \)
\( =\sqrt{34} \)
\( B D=\sqrt{(6-1)^{2}+(2-5)^{2}} \)
\( =\sqrt{(5)^{2}+(-3)^{2}} \)
\( =\sqrt{25+9} \)
\( =\sqrt{34} \)
\( \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \) 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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