We know that the square of distance between two points \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}},{y}_{{2}}\right)}{i}{s}{\left({x}_{{2}}-{x}_{{1}}\right)}^{{2}}+{\left(_{y}{2}-{y}_{{1}}\right)}^{{2}}\) and the converse of pythagorus theorem , we have if sum of square of any two side is equal to square of third side of triangle, then it is right angled triangle.

Using distance formula and converse of pythagorus theorem , we find that ABC is right angled triangle right angled at A

A(-3,1), B(2,-1), C(6,9)

\(\displaystyle{c}^{{2}}={\left|{A}{B}\right|}^{{2}}={\left({2}+{3}\right)}^{{2}}+{\left(-{1}-{1}\right)}^{{2}}={25}+{4}={29}\)

\(\displaystyle{a}^{{2}}={\left|{B}{C}\right|}^{{2}}={\left({6}-{2}\right)}^{{2}}+{\left({9}+{1}\right)}^{{2}}={16}+{100}={116}\)

\(\displaystyle{b}^{{2}}={\left|{C}{A}\right|}^{{2}}={\left({6}+{3}\right)}^{{2}}+{\left({9}-{1}\right)}^{{2}}={81}+{64}={145}\)

\(\displaystyle{a}^{{2}}\ne{b}^{{2}}+{c}^{{2}}\)

145=116+29

Using distance formula and converse of pythagorus theorem , we find that ABC is right angled triangle right angled at A

A(-3,1), B(2,-1), C(6,9)

\(\displaystyle{c}^{{2}}={\left|{A}{B}\right|}^{{2}}={\left({2}+{3}\right)}^{{2}}+{\left(-{1}-{1}\right)}^{{2}}={25}+{4}={29}\)

\(\displaystyle{a}^{{2}}={\left|{B}{C}\right|}^{{2}}={\left({6}-{2}\right)}^{{2}}+{\left({9}+{1}\right)}^{{2}}={16}+{100}={116}\)

\(\displaystyle{b}^{{2}}={\left|{C}{A}\right|}^{{2}}={\left({6}+{3}\right)}^{{2}}+{\left({9}-{1}\right)}^{{2}}={81}+{64}={145}\)

\(\displaystyle{a}^{{2}}\ne{b}^{{2}}+{c}^{{2}}\)

145=116+29