#### 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?

Solution.
Let the points be A(2, –2), B(7, 3), C(11, –1), D(6, –6) of a quadrilateral ABCD
$AB= \sqrt{\left ( 7-2 \right )^{2}+\left ( 3+2 \right )^{2}}= \sqrt{5^{2}+5^{2}}= \sqrt{25+25}= \sqrt{50}$
$BC= \sqrt{\left ( 7-11 \right )^{2}+\left ( -1-3 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}+\left ( -4 \right )^{2}}= \sqrt{16+16}= \sqrt{32}$
$CD= \sqrt{\left ( 6-11 \right )^{2}+\left ( -6+1 \right )^{2}}= \sqrt{\left ( -5 \right )^{2}+\left ( -5 \right )^{2}}= \sqrt{25+25}= \sqrt{50}$
$DA= \sqrt{\left ( 6-2 \right )^{2}+\left ( -6+2 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}+\left ( -4 \right )^{2}}= \sqrt{16+16}= \sqrt{32}$
$AC= \sqrt{\left ( 11-2 \right )^{2}+\left ( -1+2 \right )^{2}}= \sqrt{\left ( 9 \right )^{2}+\left ( 1 \right )^{2}}= \sqrt{81+1}= \sqrt{82}$
$BD= \sqrt{\left ( 6-7 \right )^{2}+\left ( -6-3 \right )^{2}}= \sqrt{\left ( 1 \right )^{2}+\left ( -9 \right )^{2}}= \sqrt{1+81}= \sqrt{82}$
As, AB = CD and BC = DA and AC = BD
Hence, the quadrilateral is a rectangle.