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Prove that the points $A (1, 7), B (4, 2), C (-1, -1)$ and $D (-4, 4)$ are the vertices of a square.
Given:
Given points are $A (1, 7), B (4, 2), C (-1, -1)$ and $D (-4, 4)$.
To do:
We have to prove that the points $A (1, 7), B (4, 2), C (-1, -1)$ and $D (-4, 4)$ 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,
\( \mathrm{AC}=\sqrt{(1+1)^{2}+(7+1)^{2}} \)
\( =\sqrt{2^{2}+8^{2}} \)
\( =\sqrt{4+64}=\sqrt{68} \)
\( \mathrm{BD}=\sqrt{(4+4)^{2}+(2-4)^{2}} \)
\( =\sqrt{(8)^{2}+(-2)^{2}} \)
\( =\sqrt{64+4}=68 \)
\( \mathrm{AC}=\mathrm{BD} \)
\( \mathrm{AB}=\sqrt{(1-4)^{2}+(7-2)^{2}} \)
\( =\sqrt{(-3)^{2}+(5)^{2}} \)
\( =\sqrt{9+25}=\sqrt{34} \)
\( \mathrm{BC}=\sqrt{(4+1)^{2}+(2+1)^{2}} \)
\( =\sqrt{(5)^{2}+(3)^{2}} \)
\( =\sqrt{25+9}=\sqrt{34} \)
\( \mathrm{CD}=\sqrt{(-1+4)^{2}+(-1-4)^{2}} \)
\( =\sqrt{(3)^{2}+(-5)^{2}} \)
\( =\sqrt{9+25}=\sqrt{34} \)
\( \mathrm{DA}=\sqrt{(1+4)^{2}+(2-4)^{2}} \)
\( =\sqrt{(5)^{2}+(3)^{2}} \)
\( =\sqrt{25+9}=\sqrt{34} \)
\( \therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \) 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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