Show that the points $ (0,0),(7,0),(7,5) $ and $ (0,5) $ are the vertices of a rectangle.

Given:

Given points are \( (0,0),(7,0),(7,5) \) and \( (0,5) \).

To do:

We have to show that the points \( (0,0),(7,0),(7,5) \) and \( (0,5) \) are the vertices of a rectangle.

Solution:

 Let \( \mathrm{ABCD} \) be a rectangle whose vertices are \( \mathrm{A}(0,0), \mathrm{B}(7,0), \mathrm{C}(7,5) \) and \( \mathrm{D}(0,5) \).

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,

\( \mathrm{AB}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \)

\( =\sqrt{(7-0)^{2}+(0-0)^{2}} \)

\( =\sqrt{(7)^{2}+(0)^{2}} \)

\( =\sqrt{49} \)

\( =7 \)

\( \mathrm{CD}=\sqrt{(0-7)^{2}+(5-5)^{2}} \)

\( =\sqrt{(-7)^{2}+(0)^{2}} \)

\( =\sqrt{49} \)

\( =7 \)

\( \mathrm{AD}=\sqrt{(0-0)^{2}+(5-0)^{2}} \)

\( =\sqrt{(0)^{2}+(5)^{2}} \)

\( =\sqrt{25} \)

\( =5 \)

\( \mathrm{BC}=\sqrt{(7-7)^{2}+(5-0)^{2}} \)

\( =\sqrt{(0)^{2}+(5)^{2}} \)

\( =\sqrt{25} \)

\( =5 \)

\( \mathrm{AC}=\sqrt{(7-0)^{2}+(5-0)^{2}} \)

\( =\sqrt{(7)^{2}+(5)^{2}} \)

\( =\sqrt{49+25} \)

\( =\sqrt{74} \)

\( \mathrm{BD}=\sqrt{(0-7)^{2}+(5-0)^{2}} \)

\( =\sqrt{(7)^{2}+(5)^{2}} \)

\( =\sqrt{49+25} \)

\( =\sqrt{74} \)

$AB = CD$ and $AD = BC$

$AC = BD$

Here, opposite sides are equal and the diagonals are equal.
Therefore, $ABCD$ is a rectangle.  

Tutorialspoint

Updated on: 10-Oct-2022

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