Question

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

Answer 1

Step1 : Apply distance formula :

The points are $A(2,–2),B(7,3),C(11,–1)$ and $D(6,–6)$

Using distance formula,

$\begin{array}{c}\mathrm{d}=\sqrt{\left({\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2\right)}\\ AB=\sqrt{(7-2{)}^2+(3+2{)}^2}=\sqrt{(5{)}^2+(5{)}^2}=\sqrt{25+25}=\sqrt{50}\\ BC=\sqrt{(11-7{)}^2+(-1-3{)}^2}=\sqrt{(4{)}^2+(-4{)}^2}=\sqrt{16+16}=4\sqrt{2}\\ CD=\sqrt{(6-11{)}^2+(-6+1{)}^2}=\sqrt{(-5{)}^2+(-5{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\\ DA=\sqrt{(2-6{)}^2+(-2+6{)}^2}=\sqrt{(-4{)}^2+(4{)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\end{array}$

Step 2: Let us determine the distance between the diagonals :

$\begin{array}{c}AC=\sqrt{(11-2{)}^2+(-1+2{)}^2}\\ =\sqrt{(9{)}^2+(1{)}^2}\\ =\sqrt{81+1}\\ =\sqrt{82}\\ BD=\sqrt{(6-7{)}^2+(-6-3{)}^2}\\ =\sqrt{(-1{)}^2+(-9{)}^2}\\ =\sqrt{1+81}\\ =\sqrt{82}\end{array}$

Hence, The Quadrilateral formed by the points $A(2,–2),B(7,3),C(11,–1)$ and $D(6,–6)$ is rectangle.