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Show that the points $ (1,-2),(2,3),(-3,2) $ and $ (-4,-3) $ are the vertices of a rhombus.
Given:
Given points are \( (1,-2),(2,3),(-3,2) \) and \( (-4,-3) \).
To do:
We have to show that the \( (1,-2),(2,3),(-3,2) \) and \( (-4,-3) \) are the vertices of a rhombus.
Solution:
Let \( \mathrm{ABCD} \) is a quadrilateral whose vertices are \( \mathrm{A}(1,-2), \mathrm{B}(2,3), \mathrm{C}(-3,2) \) and \( \mathrm{D}(-4,-3) \).
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,
\( A B=\sqrt{(2-1)^{2}+(3+2)^{2}} \)\( =\sqrt{(1)^{2}+(5)^{2}} \)
\( =\sqrt{1+25} \)
\( =\sqrt{26} \)
Similarly,
\( B C=\sqrt{(-3-2)^{2}+(2-3)^{2}} \)
\( =\sqrt{(-5)^{2}+(-1)^{2}} \)
\( =\sqrt{25+1} \)
\( =\sqrt{26} \)
\( C D=\sqrt{(-4+3)^{2}+(-3-2)^{2}} \)
\( =\sqrt{(-1)^{2}+(-5)^{2}} \)
\( =\sqrt{1+25} \)
\( =\sqrt{26} \)
\( D A=\sqrt{(-4-1)^{2}+(-3+2)^{2}} \)
\( =\sqrt{(-5)^{2}+(-1)^{2}} \)
\( =\sqrt{25+1} \)
\( =\sqrt{26} \)
Diagonal \( \mathrm{AC}=\sqrt{(-3-1)^{2}+(2+2)^{2}} \)
\( =\sqrt{(-4)^{2}+(4)^{2}} \)
\( =\sqrt{16+16} \)
\( =\sqrt{32} \)
\( =\sqrt{16 \times 2} \)
\( =4 \sqrt{2} \)
Diagonal \( \mathrm{BD}=\sqrt{(-4-2)^{2}+(-3-3)^{2}} \)
\( =\sqrt{(-6)^{2}+(-6)^{2}} \)
\( =\sqrt{36+36} \)
\( =\sqrt{72} \)
\( =\sqrt{36 \times 2} \)
\( =6 \sqrt{2} \)
Here,
$A B=B C=C D=D A=\sqrt{26}$
Sides are equal but diagonals are not equal.
Therefore, the given points are the vertices of a rhombus.