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Show that the points $(-4, -1), (-2, -4), (4, 0)$ and $(2, 3)$ are the vertices points of a rectangle.
Given:
Given points are $(-4, -1), (-2, -4), (4, 0)$ and $(2, 3)$.
To do:
We have to show that the points $(-4, -1), (-2, -4), (4, 0)$ and $(2, 3)$ are the vertices points of a rectangle.
Solution:
Let \( \mathrm{ABCD} \) be a rectangle whose vertices are \( \mathrm{A}(-4,-1), \mathrm{B}(-2,-4), \mathrm{C}(4,0) \) and \( \mathrm{D}(2,3) \).
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,
\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)
\( =\sqrt{(-2+4)^{2}+(-4+1)^{2}} \)
\( =\sqrt{(2)^{2}+(-3)^{2}} \)
\( =\sqrt{4+9} \)
\( =\sqrt{13} \)
\( \mathrm{CD}=\sqrt{(2-4)^{2}+(3-0)^{2}} \)
\( =\sqrt{(-2)^{2}+(3)^{2}} \)
\( =\sqrt{4+9} \)
\( =\sqrt{13} \)
\( \mathrm{AD}=\sqrt{(2+4)^{2}+(3+1)^{2}} \)
\( =\sqrt{(6)^{2}+(4)^{2}} \)
\( =\sqrt{36+16} \)
\( =\sqrt{52} \)
\( \mathrm{BC}=\sqrt{(4+2)^{2}+(0+4)^{2}} \)
\( =\sqrt{(6)^{2}+(4)^{2}} \)
\( =\sqrt{36+16} \)
\( =\sqrt{52} \)
\( \mathrm{AC}=\sqrt{(4+4)^{2}+(0+1)^{2}} \)
\( =\sqrt{(8)^{2}+(1)^{2}} \)
\( =\sqrt{64+1} \)
\( =\sqrt{65} \)
\( \mathrm{BD}=\sqrt{(2+2)^{2}+(3+4)^{2}} \)
\( =\sqrt{(4)^{2}+(7)^{2}} \)
\( =\sqrt{16+49} \)
\( =\sqrt{65} \)
$AB = CD$ and $AD = BC$
$AC = BD$
Here, opposite sides are equal and the diagonals are equal.
Therefore, $ABCD$ is a rectangle.
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