প্রমাণঃ
মনে করি, কোন সমতলে \(\) ত্রিভুজের শীর্ষত্রয় যথক্রমে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\)। \(A, B, C\) বিন্দু হতে \(X\) অক্ষের উপর যথাক্রমে \(AL\), \(BM\) এবং \(CN\) লম্ব আঁকি।
তাহলে, \(OL=x_{1}\)
\(OM=x_{2}\)
\(ON=x_{3}\)
\(AL=y_{1}\)
\(BM=y_{2}\)
\(CN=y_{3}\).
আবার,
\(ML=OL-OM=x_{1}-x_{2}\)
\(LN=ON-OL=x_{3}-x_{1}\)
\(MN=ON-OM=x_{3}-x_{2}\).
এখন,
\(\triangle ABC=\) ট্রাপিজিয়াম \(ABML\) এর ক্ষেত্রফল + ট্রাপিজিয়াম \(ALNC\) এর ক্ষেত্রফল -ট্রাপিজিয়াম \(BMNC\) এর ক্ষেত্রফল।
\(=\frac{1}{2}(AL+BM)\times ML+\frac{1}{2}(AL+CN)\times LN-\frac{1}{2}(BM+CN)\times MN\)
\(=\frac{1}{2}(y_{1}+y_{2})\times (x_{1}-x_{2})+\frac{1}{2}(y_{1}+y_{3})\times (x_{3}-x_{1})-\frac{1}{2}(y_{2}+y_{3})\times (x_{3}-x_{2})\)
\(=\frac{1}{2}\{(y_{1}+y_{2})(x_{1}-x_{2})+(y_{1}+y_{3})(x_{3}-x_{1})-(y_{2}+y_{3})(x_{3}-x_{2})\}\)
\(=\frac{1}{2}\{x_{1}y_{1}+x_{1}y_{2}-x_{2}y_{1}-x_{2}y_{2}+x_{3}y_{1}+x_{3}y_{3}-x_{1}y_{1}-x_{1}y_{3}-(x_{3}y_{2}+x_{3}y_{3}-x_{2}y_{2}-x_{2}y_{3})\}\)
\(=\frac{1}{2}\{x_{1}y_{1}+x_{1}y_{2}-x_{2}y_{1}-x_{2}y_{2}+x_{3}y_{1}+x_{3}y_{3}-x_{1}y_{1}-x_{1}y_{3}-x_{3}y_{2}-x_{3}y_{3}+x_{2}y_{2}+x_{2}y_{3}\}\)
\(=\frac{1}{2}\{x_{1}y_{2}-x_{2}y_{1}+x_{3}y_{1}-x_{1}y_{3}-x_{3}y_{2}+x_{2}y_{3}\}\)
\(=\frac{1}{2}\{(x_{1}y_{2}-x_{2}y_{1})+(x_{2}y_{3}-x_{3}y_{2})+(x_{3}y_{1}-x_{1}y_{3})\}\)
\(=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
আবার,
\(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(x_{1}y_{2}-x_{2}y_{1})+(x_{2}y_{3}-x_{3}y_{2})+(x_{3}y_{1}-x_{1}y_{3})\}\)
\(=\frac{1}{2}\{x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})+1(x_{2}y_{3}-x_{3}y_{2})\}\)
\(=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ y_{1} \ \ 1\\x_{2} \ \ y_{2} \ \ 1\\x_{3} \ \ y_{3} \ \ 1\end{array}\right|\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ y_{1} \ \ 1\\x_{2} \ \ y_{2} \ \ 1\\x_{3} \ \ y_{3} \ \ 1\end{array}\right|\)
আবার,
\(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(x_{1}y_{2}-x_{2}y_{1})+(x_{2}y_{3}-x_{3}y_{2})+(x_{3}y_{1}-x_{1}y_{3})\}\)
\(=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)
\(\therefore \triangle ABC=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)

প্রমাণঃ
মনে করি, কোন সমতলে \(ABCD\) চতুর্ভুজের শীর্ষচতুষ্টয় যথক্রমে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{3}, y_{3})\)। ত্রিভুজের ক্ষেত্রফল নির্ণয়ের সূত্র ব্যাবহার করে।
\(\Box ABCD=\triangle ABC+\triangle ACD\)
\(=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)+\(\frac{1}{2}\{(x_{1}y_{3}+x_{3}y_{4}+x_{4}y_{1})-(y_{1}x_{3}+y_{3}x_{4}+y_{4}x_{1})\}\)
\(=\frac{1}{2}\{x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-y_{1}x_{2}-y_{2}x_{3}-y_{3}x_{1}\)+\(x_{1}y_{3}+x_{3}y_{4}+x_{4}y_{1}-y_{1}x_{3}-y_{3}x_{4}-y_{4}x_{1}\}\)
\(=\frac{1}{2}\{x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{4}+x_{4}y_{1}-y_{1}x_{2}-y_{2}x_{3}-y_{3}x_{4}-y_{4}x_{1}\}\)
\(=\frac{1}{2}\{(x_{1}y_{2}-y_{1}x_{2})+(x_{2}y_{3}-y_{2}x_{3})+(x_{3}y_{4}-y_{3}x_{4})+(x_{4}y_{1}-y_{4}x_{1})\}\)
\(=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4}\end{array}\right|\)
\(\therefore \Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4}\end{array}\right|\)

প্রমাণঃ
\(C\) এবং \(D\) বিন্দু দুইটি \(AB\) রেখার একই পার্শ্বে অবস্থান করবে যদি,
\(\delta_{ABC}>0, \ \ \delta_{ABD}>0\) অথবা, \(\delta_{ABC}<0 , \ \ \delta_{ABD}<0 \) হয়।
তাহলে ইহা স্পষ্ট যে,
\(\delta_{ABC}\times \delta_{ABD}>0\).

প্রমাণঃ
\(C\) এবং \(D\) বিন্দু দুইটি \(AB\) রেখার বিপরীত পার্শ্বে অবস্থান করবে যদি,
\(\delta_{ABC}>0, \ \ \delta_{ABD}<0\) অথবা, \(\delta_{ABC}<0, \ \ \delta_{ABD}>0\) হয়।
তাহলে ইহা স্পষ্ট যে,
\(\delta_{ABC}\times \delta_{ABD}<0\).

প্রমাণঃ
\(AB\) এর উপর \(CN\) ও \(DM\) লম্ব হলে, \(\triangle CNE\) ও \(\triangle DME\) সদৃশ।
\(\therefore \frac{CN}{DM}=\frac{CE}{DE}=\frac{m}{n}\) ➜Note \(\because CE=m, DE=n\)
\(\Rightarrow \frac{CN}{DM}=\frac{m}{n}\)
\(\Rightarrow \frac{\frac{1}{2}AB\times CN}{\frac{1}{2}AB\times DM}=\frac{m}{n}\) ➜Note বাম পার্শের লব ও হর কে \(\frac{1}{2}AB\) দ্বারা গুণ করে।
\(\Rightarrow \frac{\triangle ABC}{\triangle ABD}=\frac{m}{n}\)
\(\Rightarrow \frac{\frac{1}{2}\delta_{ABC}}{\frac{1}{2}\delta_{ABD}}=\frac{m}{n}\) ➜Note \(\because \triangle ABC=\frac{1}{2}\delta_{ABC}, \triangle ABD=\frac{1}{2}\delta_{ABD}\)
\(\Rightarrow \frac{\delta_{ABC}}{\delta_{ABD}}=\frac{m}{n}\) ➜Note বাম পার্শের লব ও হর কে \(2\) দ্বারা গুণ করে।
\(\therefore \delta_{ABC}:\delta_{ABD}=m:n\)

প্রমাণঃ
কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থান করবে বা সমরেখ হবে যদি, \(\triangle ABC=0\) হয়।
\(\Rightarrow \frac{1}{2}\mid \delta_{ABC} \mid=0\) ➜Note \(\because \triangle ABC= \frac{1}{2}\mid \delta_{ABC} \mid\)
\(\Rightarrow \mid \delta_{ABC} \mid=0\)
\(\Rightarrow \delta_{ABC}=0\)
\(\Rightarrow \left|\begin{array}{c}x_{1}&x_{2}&x_{3}&x_{1}\\ y_{1}& y_{2} &y_{3}&y_{1}\end{array}\right|=0\)
\(\Rightarrow x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}=0\)
\(\Rightarrow x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})+x_{2}y_{3}-x_{3}y_{2}=0\)
\(\Rightarrow x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})-x_{2}y_{2}+x_{2}y_{3}+x_{2}y_{2}-x_{3}y_{2}=0\)
\(\Rightarrow x_{1}(y_{2}-y_{3})-y_{1}(x_{2}-x_{3})-x_{2}(y_{2}-y_{3})+y_{2}(x_{2}-x_{3})=0\)
\(\Rightarrow (x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})=0\)
\(\therefore \delta_{ABC}=(x_{1}-x_{2})(y_{2}-y_{3})-(y_{1}-y_{2})(x_{2}-x_{3})=0\)