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Show that the quadrilateral whose vertices are $(2, -1), (3, 4), (-2, 3)$ and $(-3, -2)$ is a rhombus.
Given:
Given points are $(2, -1), (3, 4), (-2, 3)$ and $(-3, -2)$.
To do:
We have to show that the quadrilateral whose vertices are $(2, -1), (3, 4), (-2, 3)$ and $(-3, -2)$ is a rhombus.
Solution:
Let \( \mathrm{ABCD} \) is a quadrilateral whose vertices are \( \mathrm{A}(2,-1), \mathrm{B}(3,4), \mathrm{C}(-2,3) \) and \( \mathrm{D}(-3,-2) \)
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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(3-2)^{2}+(4+1)^{2}} \)
\( =\sqrt{(1)^{2}+(5)^{2}} \)
\( =\sqrt{1+25}=\sqrt{26} \)
\( \mathrm{BC}=\sqrt{(-2-3)^{2}+(3-4)^{2}} \)
\( =\sqrt{(-5)^{2}+(-1)^{2}} \)
\( =\sqrt{25+1}=\sqrt{26} \)
\( \mathrm{CD}=\sqrt{(-3+2)^{2}+(-2-3)^{2}} \)
\( =\sqrt{(-1)^{2}+(-5)^{2}} \)
\( =\sqrt{1+25}=\sqrt{26} \)
\( \mathrm{DA}=\sqrt{(2+3)^{2}+(-1+2)^{2}} \)
\( =\sqrt{(5)^{2}+(1)^{2}} \)
\( =\sqrt{25+1}=\sqrt{26} \)
\( \mathrm{AC}=\sqrt{(-2-2)^{2}+(3+1)^{2}} \)
\( =\sqrt{(-4)^{2}+(4)^{2}} \)
\( =\sqrt{16+16}=\sqrt{32} \)
\( \mathrm{BD}=\sqrt{(-3-3)^{2}+(-2-4)^{2}} \)
\( =\sqrt{(-6)^{2}+(-6)^{2}} \)
\( =\sqrt{36+36}=\sqrt{72} \)
\( A B=B C=C D=D A=\sqrt{26} \) and $AC≠BD$
Here, all the sides are equal and the diagonals are not equal to each other.
Therefore, the given points are the vertices of a rhombus.
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