Solution:

A (2, –2), B (7, 3), C (11, –1) and D (6, –6) are the given points.

Now using the distance formula,

$d~=\text{ }\surd \text{ }({{({{x}_{2}}~\text{ }{{x}_{1}})}^{2}}~+\text{ }{{({{y}_{2}}~\text{ }{{y}_{1}})}^{2}})$

$AB=\sqrt{{{\left( 7-2 \right)}^{2}}+{{\left( 3+2 \right)}^{2}}}$

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

$=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$

$BC=\sqrt{{{\left( 11-7 \right)}^{2}}+{{\left( -1-3 \right)}^{2}}}$

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

$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

$CD=\sqrt{{{\left( 6-11 \right)}^{2}}+{{\left( -6+1 \right)}^{2}}}$

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

$=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$

$DA=\sqrt{{{\left( 2-6 \right)}^{2}}+{{\left( -2+6 \right)}^{2}}}$

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

$=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$

Now finding the diagonals AC and BD,

$AC=\sqrt{{{\left( 11-2 \right)}^{2}}+{{\left( -1+2 \right)}^{2}}}$

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

$=\sqrt{81+1}=\sqrt{82}$

$BD=\sqrt{{{\left( 6-7 \right)}^{2}}+{{\left( -6-3 \right)}^{2}}}$

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

$=\sqrt{1+81}=\sqrt{82}$

As a result, the quadrilateral so formed is rectangle.