આપેલ રેખાના ઢાળ અનુક્રમે $ – 1\,\,,\,\, – 3\,,\,\, – \frac{1}{3}$ છે.

ધારો કે $\,{m_1}\, = \, – \frac{1}{3}\,,\,\,{m_2}\, = \,\, – 1\,,\,\,{m_3}\,\, = \,\, – 3$

$\therefore \,\,\,\tan A\,\, = \,\frac{{ – \frac{1}{3}\,\, + \,1}}{{1 + \frac{1}{3}\,1}}\,\,$$ \Rightarrow \,\,\,A\,\, = \,\,{\tan ^{ – 1}}\,\left( {\frac{1}{2}} \right)$

$\tan B\,\, = \,\,\,\,\frac{{ – 1 + 3}}{{1 + 1.3}}\,\,$ $ \Rightarrow \,\,B\,\, = \,\,{\tan ^{ – 1}}\,\left( {\frac{1}{2}} \right)\,\,$ અને $tan\,\,C\,\, = \,\, \frac{{ – 3 + \frac{1}{3}}}{{1 + 3\frac{1}{3}}}\,\,$ $\, \Rightarrow \,\,\,C\,\, = \,\,\,{\tan ^{ – 1}}\,\left( { – \frac{4}{3}} \right)$

∵ $\angle A = \angle B,$ આથી ત્રિકોણ સમદ્રિબાજુ ત્રિકોણ છે.