মনে করি,
কোন সমতলের উপর \(P(r_{1}, \theta_{1})\) ও \(Q(r_{2}, \theta_{2})\) যে কোন দুইটি বিন্দু এদের মধ্যবর্তী দূরত্ব নির্ণয় করতে হবে।
\(P(r_{1}, \theta_{1})\) ও \(Q(r_{2}, \theta_{2})\) বিন্দুদ্বয়ের কার্তেসীয় স্থানাঙ্ক যথাক্রমে \(P(r_{1}\cos\theta_{1}, r_{1}\sin\theta_{1})\) ও \(Q(r_{2}\cos\theta_{2}, r_{2}\sin\theta_{2})\)
\(\therefore PQ=\sqrt{(r_{1}\cos\theta_{1}-r_{2}\cos\theta_{2})^{2}+(r_{1}\sin\theta_{1}-r_{2}\sin\theta_{2})^{2}}\)
\(\Rightarrow PQ=\sqrt{r_{1}^{2}\cos^{2}\theta_{1}-2r_{1}r_{2}\cos\theta_{1}\cos\theta_{2}+r_{2}^{2}\cos^{2}\theta_{2}+r_{1}^{2}\sin^{2}\theta_{1}-2r_{1}r_{2}\sin\theta_{1}\sin\theta_{2}+r_{2}^{2}\sin^{2}\theta_{2}}\)
\(\Rightarrow PQ=\sqrt{r_{1}^{2}(\cos^{2}\theta_{1}+\sin^{2}\theta_{1})+r_{2}^{2}(\cos^{2}\theta_{2}+\sin^{2}\theta_{2})-2r_{1}r_{2}(\cos\theta_{1}\cos\theta_{2}+\sin\theta_{1}\sin\theta_{2})}\)
\(\Rightarrow PQ=\sqrt{r_{1}^{2}\times 1+r_{2}^{2}\times 1-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}\)
\(\Rightarrow PQ=\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}\)

মনে করি,
কোন সমতলের উপর তিনটি বিন্দু \(P(x_{1}, y_{1}), \ Q(x_{2}, y_{2})\) এবং \(R(x_{3}, y_{3})\)
এখন, \(\triangle{PQR}=\frac{1}{2}\left|\begin{array}{c}x_{1}&x_{2}&x_{3}&x_{1}\\ y_{1}&y_{2}&y_{3}&y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}\}\)
\(=\frac{1}{2}\{x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})+x_{2}y_{3}-x_{3}y_{2}\}\)
\(=\frac{1}{2}\{x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})-x_{2}y_{2}+x_{2}y_{3}+x_{2}y_{2}-x_{3}y_{2}\}\)
\(=\frac{1}{2}\{x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})-x_{2}(y_{2}-y_{3})+y_{2}(x_{2}-x_{3})\}\)
\(=\frac{1}{2}\{(x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})\}\)
\(\therefore \triangle{PQR}=\frac{1}{2}\{(x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})\}\)
বিন্দু তিনটি সমরেখ হবে যদি,
\(\triangle{PQR}=0\) হয়।
\(\Rightarrow \frac{1}{2}\{(x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})\}=0\)
\(\therefore (x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})=0\)
(প্রমাণিত)