# 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 CAC=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.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

9 Views