What type of a quadrilateral do the points $ A(2,-2), B(7,3), C(11,-1) $ and $ \mathrm{D}(6,-6) $ taken in that order, form?

Given:

The points \( A(2,-2), B(7,3), C(11,-1) \) and \( \mathrm{D}(6,-6) \).

To do:

We have to find the type of quadrilateral formed by the given points.

Solution:

The distance between the points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

The distance between $A(2,-2)$ and $B(7,3) is,

$A B=\sqrt{(7-2)^{2}+(3+2)^{2}}$

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

$=\sqrt{25+25}$

$=\sqrt{50}$

$=5 \sqrt{2}$

The distance between $B(7,3)$ and $C(11,-1)$ is,

$B C=\sqrt{(11-7)^{2}+(-1-3)^{2}}$

$=\sqrt{(4)^{2}+(-4)^2}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4 \sqrt{2}$

The distance between $C(11,-1)$ and $D(6,-6)$ is,

$C D=\sqrt{(6-11)^{2}+(-6+1)^{2}}$

$=\sqrt{(-5)^{2}+(-5)^{2}}$

$=\sqrt{25+25}$

$=\sqrt{50}$

$=5 \sqrt{2}$

The distance between $D(6,-6)$ and $C(2,-2)$ is,

$D A=\sqrt{(2-6)^{2}+(-2+6)^{2}}$

$=\sqrt{(-4)^{2}+(4)^{2}}$

$=\sqrt{16+16}$

$=\sqrt{32}$

$=4 \sqrt{2}$

The distance between $A(2,-2)$ and $C(11,-1)$ is,

$A C=\sqrt{(11-2)^{2}+(-1+2)^{2}}$

$=\sqrt{(9)^{2}+(1)^{2}}$

$=\sqrt{81+1}$

$=\sqrt{82}$

The distance between $D(6,-6)$ and $B(7,3)$ is,

$B D=\sqrt{(6-7)^{2}+(-6-3)^{2}}$

$=\sqrt{(-1)^{2}+(-9)^{2}}$

$=\sqrt{1+81}$

$=\sqrt{82}$

We know that, in a rectangle, opposite sides are equal and diagonals are equal to each other.

Here,

$A B=C D$ and $A D=B C$

$AC=D$

Therefore, the points \( A(2,-2), B(7,3), C(11,-1) \) and \( \mathrm{D}(6,-6) \) taken in that order form a rectangle.

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Updated on: 10-Oct-2022

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