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 \(MBNC\) 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,
\(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(x_{1}y_{2}-x_{2}y_{1})+(x_{2}y_{3}-x_{3}y_{2})+(x_{3}y_{1}-x_{1}y_{3})\}\)
\(=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)
\(\therefore \triangle ABC=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)

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।
\(\Box ABCD=\triangle ABC+\triangle ACD\)
\(=\frac{1}{2}\{(x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1})-(y_{1}x_{2}+y_{2}x_{3}+y_{3}x_{1})\}\)+\(\frac{1}{2}\{(x_{1}y_{3}+x_{3}y_{4}+x_{4}y_{1})-(y_{1}x_{3}+y_{3}x_{4}+y_{4}x_{1})\}\)
\(=\frac{1}{2}\{x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{1}-y_{1}x_{2}-y_{2}x_{3}-y_{3}x_{1}\)+\(x_{1}y_{3}+x_{3}y_{4}+x_{4}y_{1}-y_{1}x_{3}-y_{3}x_{4}-y_{4}x_{1}\}\)
\(=\frac{1}{2}\{x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{4}+x_{4}y_{1}-y_{1}x_{2}-y_{2}x_{3}-y_{3}x_{4}-y_{4}x_{1}\}\)
\(=\frac{1}{2}\{(x_{1}y_{2}-y_{1}x_{2})+(x_{2}y_{3}-y_{2}x_{3})+(x_{3}y_{4}-y_{3}x_{4})+(x_{4}y_{1}-y_{4}x_{1})\}\)
\(=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4}\end{array}\right|\)
\(\therefore \Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4}\end{array}\right|\)

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

\(C\) এবং \(D\) বিন্দু দুইটি \(AB\) রেখার বিপরীত পার্শে 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।
\(\therefore \delta_{ABC}:\delta_{ABD}=m:n\)

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 \(\because \triangle ABC= \frac{1}{2}\mid \delta_{ABC} \mid\)
\(\Rightarrow \mid \delta_{ABC} \mid=0\)
\(\therefore \delta_{ABC}=0\)

মনে করি, \(A(6, 5)\), \(B(-9, -4)\) এবং \(C(-5, 0)\).
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}6 \ \ -9 \ \ -5 \ \ \ \ 6\\ 5 \ \ -4 \ \ \ \ \ 0 \ \ \ \ \ 5\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{6.(-4)-5.(-9)+(-9).0-(-5).(-4)+(-5).5-0.6\}\)
\(=\frac{1}{2}\{-24+45-0-20-25-0\}\)
\(=\frac{1}{2}\{45-69\}\)
\(=\frac{1}{2}\times -24\)
\(=-12\)
\(=12\) ➜Note \(\because \triangle \neq -ve\)
\(\therefore \triangle ABC=12\) বর্গ একক।

মনে করি, \(A(a, 2-2a)\), \(B(1-a, 2a)\) এবং \(C(-4-a, 6-2a)\) বিন্দুতিনটি সমরেখ।
\(\therefore \delta_{ABC}=0\)
\(\Rightarrow \left|\begin{array}{c}a \ \ \ \ 1-a \ \ -4-a \ \ \ \ a\\ 2-2a \ \ 2a \ \ 6-2a \ \ 2-2a\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \{a.2a-(1-a)(2-2a)+(1-a)(6-2a)-(-4-a).2a+(-4-a)(2-2a)-a(6-2a)\}=0\)
\(\Rightarrow 2a^{2}-2+2a+2a-2a^{2}+6-6a-2a+2a^{2}+8a+2a^{2}-8-2a+8a+2a^{2}-6a+2a^{2}=0\)
\(\Rightarrow 8a^{2}+4a-4=0\)
\(\Rightarrow 2a^{2}+a-1=0\) ➜Note উভয় পার্শে \(4\) ভাগ করে।
\(\Rightarrow 2a^{2}+2a-a-1=0\)
\(\Rightarrow 2a(a+1)-1(a+1)=0\)
\(\Rightarrow (a+1)(2a-1)=0\)
\(\Rightarrow a+1=0, 2a-1=0\)
\(\Rightarrow a=-1, 2a=1\)
\(\Rightarrow a=-1, a=\frac{1}{2}\)
\(\therefore a=-1, \frac{1}{2}\).

দেওয়া আছে, \(A(x, y)\), \(B(1, 2)\) এবং \(C(2, 1)\)।
শর্তমতে, \(\triangle ABC=\pm 6\)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{c}x \ \ 1 \ \ 2 \ \ x\\ y \ \ 2 \ \ 1 \ \ y\end{array}\right|=\pm 6\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \frac{1}{2}\{(2x-y)+(1-4)+(2y-x)\}=\pm6\)
\(\Rightarrow \frac{1}{2}\{2x-y+1-4+2y-x\}=\pm 6\)
\(\Rightarrow 2x-y+1-4+2y-x=\pm 12\)
\(\Rightarrow x+y-3=\pm 12\)
\(\Rightarrow x+y=\pm 12+3\)
\(\Rightarrow x+y=+12+3\) ➜Note ধনাত্মক চিহ্ন \(+ve\) ব্যাবহার করে।
\(\therefore x+y=15\)
আবার,
\(\Rightarrow x+y=-12+3\) ➜Note ঋনাত্মক চিহ্ন \(-ve\) ব্যাবহার করে।
\(\Rightarrow x+y=-9\)
\(\therefore x+y+9=0\)
[দেখানো হলো।]

দেওয়া আছে, \(A(-3, -2)\), \(B(-3, 9)\) এবং \(C(5, -8)\)।
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}-3 \ \ -3 \ \ \ \ 5 \ \ -3\\ -2 \ \ \ \ 9 \ \ -8 \ \ -2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-3).9-(-3).(-2)+(-3).(-8)-5.9+5.(-2)-(-3)(-8)\}\)
\(=\frac{1}{2}\{-27-6+24-45-10-24\}\)
\(=\frac{1}{2}\times -88\)
\(=-44\)
\(=44\) ➜Note \(\because \triangle \neq -ve\)
\(\therefore \triangle ABC=44\) বর্গ একক।
এখন,
\(CA=\sqrt{(5+3)^{2}+(-8+2)^{2}}\)
\(=\sqrt{8^{2}+(-6)^{2}}\)
\(CA=\sqrt{64+36}\)
\(CA=\sqrt{100}\)
\(CA=\sqrt{10^{2}}\)
\(\therefore CA=10\)
\(B\) বিন্দু হতে \(CA\) বাহুর উপর অঙ্কিত লম্বের দৈর্ঘ্য \(BD\)
\(\therefore \frac{1}{2}\times CA\times BD=\triangle ABC\)
\(\Rightarrow \frac{1}{2}\times 10\times BD=44\)
\(\Rightarrow 5\times BD=44\)
\(\therefore BD=\frac{44}{5}\) একক।

দেওয়া আছে, \(A(5, 6)\), \(B(-9, 1)\) এবং \(C(-3, -1)\)।
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}5 \ \ -9 \ \ -3 \ \ \ \ \ 5\\6 \ \ \ \ \ 1 \ \ -1 \ \ \ \ \ 6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{5.1-(-9).6+(-9).(-1)-(-3).1+(-3).6-5.(-1)\}\)
\(=\frac{1}{2}\{5+54+9+3-18+5\}\)
\(=\frac{1}{2}\{76-18\}\)
\(=\frac{1}{2}\times 58\)
\(=29\)
\(\therefore \triangle ABC=29\) বর্গ একক।
এখন,
\(BC=\sqrt{(-9+3)^{2}+(1+1)^{2}}\)
\(=\sqrt{6^{2}+2^{2}}\)
\(CA=\sqrt{36+4}\)
\(CA=\sqrt{10\times 4}\)
\(CA=\sqrt{10\times 2^{2}}\)
\(\therefore CA=2\sqrt{10}\)
\(A\) বিন্দু হতে \(BC\) বাহুর উপর অঙ্কিত লম্বের দৈর্ঘ্য \(AD\)
\(\therefore \frac{1}{2}\times BC\times AD=\triangle ABC\)
\(\Rightarrow \frac{1}{2}\times 2\sqrt{10}\times AD=29\)
\(\Rightarrow \sqrt{10} AD=29\)
\(\Rightarrow AD=\frac{29}{\sqrt{10}}\)
\(\therefore AD=\frac{29\times \sqrt{10}}{(\sqrt{10})^{2}}\)
\(\therefore AD=\frac{29}{10}\sqrt{10}\) একক।

দেওয়া আছে, \(A(3, 4)\), \(B(2t, 5)\) এবং \(C(6, t)\)।
শর্তমতে,
\(\triangle ABC=+19\frac{1}{2}\)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{c}3 \ \ 2t \ \ 6 \ \ 3\\ 4 \ \ 5 \ \ t \ \ 4\end{array}\right|=+19\frac{1}{2}\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \left|\begin{array}{c}3 \ \ 2t \ \ 6 \ \ 3\\ 4 \ \ 5 \ \ t \ \ 4\end{array}\right|=+\frac{39}{2}\times 2\)
\(\Rightarrow 3.5-2t.4+2t.t-5.6+6.4-3.t=39\)
\(\Rightarrow 15-8t+2t^{2}-30+24-3t-39=0\)
\(\Rightarrow 2t^{2}-11t-30+39-39=0\)
\(\Rightarrow 2t^{2}-11t-30=0\)
\(\Rightarrow 2t^{2}-15t+4t-30=0\)
\(\Rightarrow t(2t-15)+2(2t-15)=0\)
\(\Rightarrow (2t-15)(t+2)=0\)
\(\Rightarrow 2t-15=0, \ t+2=0\)
\(\Rightarrow 2t=15, \ t=-2\)
\(\Rightarrow t=\frac{15}{2}, \ t=-2\)
\(\therefore t=\frac{15}{2}, -2\)

দেওয়া আছে, \(A(3, 2)\), \(B(-2, 3)\), \(C(-1, -1)\) এবং \(D(2, -1)\)।
\(\therefore \triangle ABCD=\frac{1}{2}\left|\begin{array}{c}3 \ \ -2 \ \ -1 \ \ \ \ \ 2 \ \ \ \ 3\\2 \ \ \ \ \ 3 \ \ -1 \ \ -1 \ \ \ \ 2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দুচারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{3.3-(-2).2+(-2)(-1)-(-1).3+(-1)(-1)-2.(-1)+2.2-3.(-1)\}\)
\(=\frac{1}{2}\{9+4+2+3+1+2+4+3\}\)
\(=\frac{1}{2}\{28\}\)
\(=14\)
\(\therefore \triangle ABCD=14\) বর্গ একক।

দেওয়া আছে, \(A(3, 1)\), \(B(1, 0)\), \(C(5, 1)\) এবং \(D(-10, -4)\)।
মনে করি, \(CD\) সরলরেখা \(AB\) সরলরেখাকে \(P(x, y)\) বিন্দুতে \(m:n\) অনুপাতে বহিঃস্থভাবে বিভক্ত করে।
\(\therefore P(\frac{m\times 1-n\times 3}{m-n}, \frac{m\times 0-n\times 1}{m-n})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে বহিঃর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2-nx_1}{m-n}, \frac{my_2-ny_1}{m-n}\right)\)
\(\Rightarrow P(\frac{m-3n}{m-n}, \frac{-n}{m-n})\)
এখন,
\(C, D, P\) একই সরলরেখায় অবস্থিত।
\(\therefore \delta_{CDP}=0\)
\(\Rightarrow \left|\begin{array}{c}5 \ \ -10 \ \ \frac{m-3n}{m-n} \ \ \ \ 5\\ 1 \ \ -4 \ \ \frac{-n}{m-n} \ \ \ \ 1\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(\Rightarrow 5.(-4)-(-10).1+(-10).(\frac{-n}{m-n})-(4).(\frac{m-3n}{m-n})+\frac{m-3n}{m-n}.1-5.(\frac{-n}{m-n})=0\)
\(\Rightarrow -20+10+\frac{10n}{m-n}+\frac{4m-12n}{m-n}+\frac{m-3n}{m-n}+\frac{5n}{m-n}=0\)
\(\Rightarrow -10+\frac{10n}{m-n}+\frac{4m-12n}{m-n}+\frac{m-3n}{m-n}+\frac{5n}{m-n}=0\)
\(\Rightarrow -10m+10n+10n+4m-12n+m-3n+5n=0\) ➜Note উভয়পার্শে \((m-n)\) গুণ করে।
\(\Rightarrow -5m+10n=0\)
\(\Rightarrow -5m=-10n\)
\(\Rightarrow m=2n\) ➜Note উভয়পার্শে \((-5)\) ভাগ করে।
\(\Rightarrow \frac{m}{n}=\frac{2}{1}\)
\(\Rightarrow m:n=2:1\)
\(\therefore CD\) সরলরেখা \(AB\) সরলরেখাকে \(2:1\) অনুপাতে বহির্বিভক্ত করে।

দেওয়া আছে, \(A(3, 1)\), \(B(1, 0)\), \(C(5, 1)\) এবং \(D(-10, -4)\) এবং \(CD\) সরলরেখা \(AB\) সরলরেখাকে বিভক্ত করে।
এখন,
\(\delta_{ACD}=\left|\begin{array}{c}3 \ \ \ \ 5 \ \ -10 \ \ \ \ 3\\ 1 \ \ \ \ 1 \ \ -4 \ \ \ \ 1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(=3.1-5.1+5.(-4)-(-10).1+(-10).1-3.(-4)\)
\(=3-5-20+10-10+12\)
\(=25-35\)
\(=-10\)
আবার,
\(\delta_{BCD}=\left|\begin{array}{c}1 \ \ \ \ 5 \ \ -10 \ \ \ \ 1\\ 0 \ \ \ \ 1 \ \ -4 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(=1.1-5.0+5.(-4)-(-10).1+(-10).0-1.(-4)\)
\(=1-0-20+10+0+4\)
\(=15-20\)
\(=-5\)
আবার,
\(\therefore \frac{\delta_{ACD}}{\delta_{BCD}}=\frac{-10}{-5}\)
\(\Rightarrow \frac{\delta_{ACD}}{\delta_{BCD}}=\frac{2}{1}\)
\(\Rightarrow \delta_{ACD}:\delta_{BCD}=2:1\)
\(\therefore CD\) সরলরেখা \(AB\) সরলরেখাকে \(2:1\) অনুপাতে বহির্বিভক্ত করে। ➜Note \(\because\) অনুপাত ধনাত্মক \((+ve)\)

মনে করি, \(A(1, 0)\), \(B(2, 1)\) এবং \(B(4, 5)\).
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}1 \ \ \ \ 2 \ \ \ \ 4 \ \ \ \ 1\\ 0 \ \ \ \ 1 \ \ \ \ 5 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{1-0+10-4+0-5\}\)
\(=\frac{1}{2}\{11-9\}\)
\(=\frac{1}{2}\times 2\)
\(=1\) বর্গ একক।

মনে করি, \(A(0, 0)\), \(B(3, 3)\) এবং \(B(3, -5)\).
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}0 \ \ \ \ \ 3 \ \ \ \ 3 \ \ \ \ \ 0\\ 0 \ \ \ 3 \ \ -5 \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.3-3.0+3.(-5)-3.3+3.0-0.(-5)\}\)
\(=\frac{1}{2}\{0-0-15-9+0+0\}\)
\(=\frac{1}{2}\{-15-9\}\)
\(=\frac{1}{2}\times -24\)
\(=-12\)
\(\therefore \triangle ABC=12\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)

মনে করি, \(A(-4, 3)\), \(B(-1, -2)\) এবং \(C(3, -2)\).
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}-4 \ \ -1 \ \ \ \ 3 \ \ -4\\ 3 \ \ -2 \ \ -2 \ \ \ 3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-4).(-2)-(-1).3+(-1).(-2)-3.(-2)+3.3-(-4).(-2)\}\)
\(=\frac{1}{2}\{8+3+2+6+9-8\}\)
\(=\frac{1}{2}\times 20\)
\(=10\)
\(\therefore \triangle ABC=10\) বর্গ একক।

মনে করি, \(A(a, a^{2})\), \(B(b, b^{2})\) এবং \(C(c, c^{2})\).
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}a \ \ \ \ b \ \ \ \ c \ \ \ \ a\\ a^{2} \ \ b^{2} \ \ c^{2} \ \ a^{2}\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{a.b^{2}-b.a^{2}+b.c^{2}-c.b^{2}+c.a^{2}-a.c^{2}\}\)
\(=\frac{1}{2}\{ab^{2}-a^{2}b+bc^{2}-b^{2}c+ca^{2}-c^{2}a\}\)
\(=\frac{1}{2}\{c(a^{2}-b^{2})-ab(a-b)-c^{2}(a-b)\}\)
\(=\frac{1}{2}(a-b)\{c(a+b)-ab-c^{2}\}\)
\(=\frac{1}{2}(a-b)\{ca+bc-ab-c^{2}\}\)
\(=\frac{1}{2}(a-b)\{c(b-c)-a(b-c)\}\)
\(=\frac{1}{2}(a-b)(b-c)\{c-a\}\)
\(=\frac{1}{2}(a-b)(b-c)(c-a)\)
\(\therefore \triangle ABC=\frac{1}{2}(a-b)(b-c)(c-a)\) বর্গ একক।

মনে করি, \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(O(0, 0)\).
\(A\) ও \(B\) বিন্দু হতে \(OX\) এর উপর যথাক্রমে \(AL\) ও \(BM\) লম্ব আঁকি।
এখানে,
\(OL=x_{1}, OM=x_{2} \therefore LM=OM-OL=x_{2}-x_{1}\)
\(AL=y_{1}, BM=y_{2}\)
এখন,
\(\triangle OAB=\triangle AOL+\)ট্রাপিজিয়াম \(ALMB\) -\(\triangle OBM\)
\(=\frac{1}{2}\times OL\times AL+\)\(\frac{AL+BM}{2}\times LM\) -\(\frac{1}{2}\times BM\times OM\)
\(=\frac{1}{2}\times x_{1}\times y_{1}+\)\(\frac{y_{1}+y_{2}}{2}\times (x_{2}-x_{1})\) -\(\frac{1}{2}\times y_{2}\times x_{2}\)
\(=\frac{1}{2}\{x_{1}y_{1}+(y_{1}+y_{2})(x_{2}-x_{1})-x_{2}y_{2}\}\)
\(=\frac{1}{2}(x_{1}y_{1}+x_{2}y_{1}+x_{2}y_{2}-x_{1}y_{1}-x_{1}y_{2}-x_{2}y_{2})\)
\(=\frac{1}{2}(x_{2}y_{1}-x_{1}y_{2})\)
\(=-\frac{1}{2}(x_{1}y_{2}-x_{2}y_{1})\)
\(=\frac{1}{2}(x_{1}y_{2}-x_{2}y_{1})\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)

দেওয়া আছে, \(A(-3, -2)\), \(B(-3, 9)\) এবং \((5, -8)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}-3 \ \ -3 \ \ \ \ 5 \ \ -3\\ -2 \ \ \ \ 9 \ \ -8 \ \ -2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-3).9-(-3).(-2)+(-3).(-8)-5.9+5.(-2)-(-3).(-8)\}\)
\(=\frac{1}{2}\{-27-6+24-45-10-24\}\)
\(=\frac{1}{2}\times -88\)
\(=-44\)
\(\therefore \triangle ABC=44\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)

দেওয়া আছে, \(A(a, bc)\), \(B(b, ca)\) এবং \((c, ab)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}a \ \ \ \ b \ \ \ \ c \ \ \ \ a\\ bc \ \ ca \ \ ab \ \ bc\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{a.ca-b.bc+b.ab-c.ca+c.bc-a.ab\}\)
\(=\frac{1}{2}\{a^{2}c-b^{2}c+ab^{2}-ac^{2}+bc^{2}-a^{2}b\}\)
\(=\frac{1}{2}\{c(a^{2}-b^{2})-ab(a-b)-c^{2}(a-b)\}\)
\(=\frac{1}{2}(a-b)\{c(a+b)-ab-c^{2}\}\)
\(=\frac{1}{2}(a-b)\{ca+bc-ab-c^{2}\}\)
\(=\frac{1}{2}(a-b)\{c(b-c)-a(b-c)\}\)
\(=\frac{1}{2}(a-b)(b-c)\{c-a\}\)
\(=\frac{1}{2}(a-b)(b-c)(c-a)\)
\(\therefore \triangle ABC=\frac{1}{2}(a-b)(b-c)(c-a)\) বর্গ একক।

দেওয়া আছে, \(A(a, b+c)\), \(B(b, c+a)\) এবং \((c, a+b)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}a \ \ \ \ \ \ \ \ b \ \ \ \ \ \ \ \ c \ \ \ \ \ \ \ \ a\\ b+c \ c+a \ a+b \ b+c\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{a.(c+a)-b.(b+c)+b.(a+b)-c.(c+a)+c.(b+c)-a.(a+b)\}\)
\(=\frac{1}{2}\{ac+a^{2}-b^{2}-bc+ab+b^{2}-c^{2}-ac+bc+c^{2}-a^{2}-ab\}\)
\(=\frac{1}{2}\times 0\)
\(=0\)
\(\therefore \triangle ABC=0\) বর্গ একক।

মনে করি, \(BC\), \(CA\) এবং \(AB\) এর মধ্যবিন্দুগুলি যথাক্রমে \(D(1, 2)\), \(E(4, 4)\) এবং \(F(2, 8)\)
\(\therefore \triangle DEF=\frac{1}{2}\left|\begin{array}{c}1 \ \ \ \ 4 \ \ \ \ 2 \ \ \ \ 1\\ 2 \ \ \ \ 4 \ \ \ \ 8 \ \ \ \ 2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{1.4-4.2+4.8-2.4+2.2-1.8\}\)
\(=\frac{1}{2}\{4-8+32-8+4-8\}\)
\(=\frac{1}{2}\{40-24\}\)
\(=\frac{1}{2}\times 16\)
\(=8\)
\(\therefore \triangle DEF=8\) বর্গ একক।
এখন,
\(\triangle ABC=4\times \triangle DEF\)
\(=4\times 8\)
\(=32\) বর্গ একক।

মনে করি,\(A(3, 5)\), \(B(3, 8)\) এবং মূলবিন্দু \(O(0, 0)\)
\(\therefore \triangle OAB=\frac{1}{2}\left|\begin{array}{c}0 \ \ \ \ 3 \ \ \ \ 3 \ \ \ \ 0\\ 0 \ \ \ \ 5 \ \ \ \ 8 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.5-3.0+3.8-3.5+3.0-0.8\}\)
\(=\frac{1}{2}\{0-0+24-15+0-0\}\)
\(=\frac{1}{2}\{24-15\}\)
\(=\frac{1}{2}\times 9\)
\(=\frac{9}{2}\)
\(\therefore \triangle DEF=4\frac{1}{2}\) বর্গ একক।

মনে করি,\(\triangle ABC\) এর মধ্যমাগুলি \(AD\), \(BE\) এবং \(CF\) এর মধ্যবিন্দুগুলি যথাক্রমে \(P(1, 2)\), \(Q(4, 4)\) এবং \(R(2, 8)\)
\(\therefore \triangle PQR=\frac{1}{2}\left|\begin{array}{c}1 \ \ \ \ 4 \ \ \ \ 2 \ \ \ \ 1\\ 2 \ \ \ \ 4 \ \ \ \ 8 \ \ \ \ 2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{1.4-4.2+4.8-2.4+2.2-1.8\}\)
\(=\frac{1}{2}\{4-8+32-8+4-8\}\)
\(=\frac{1}{2}\{40-24\}\)
\(=\frac{1}{2}\times 16\)
\(=8\)
\(\therefore \triangle PQR=8\) বর্গ একক।
এখন,
\(\triangle ABC=16\times \triangle PQR\)
\(=16\times 8\)
\(=128\) বর্গ একক।

মনে করি, \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(P(x_{1}+h, y_{1}+k)\), \(Q(x_{2}+h, y_{2}+k)\), \(R(x_{3}+h, y_{3}+k)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ \ \ x_{2} \ \ \ \ x_{3} \ \ \ \ x_{1}\\ y_{1} \ \ \ \ y_{2} \ \ \ \ y_{3} \ \ \ \ y_{1}\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}\}\)
\(\therefore \triangle ABC=\frac{1}{2}\{x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}\}\) বর্গ একক।
আবার,
\(\triangle PQR=\frac{1}{2}\left|\begin{array}{c}x_{1}+h \ \ \ \ x_{2}+h \ \ \ \ x_{3}+h \ \ \ \ x_{1}+h\\ y_{1}+k \ \ \ \ y_{2}+k \ \ \ \ y_{3}+k \ \ \ \ y_{1}+k\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(x_{1}+h)(y_{2}+k)-(x_{2}+h)(y_{1}+k)+(x_{2}+h)(y_{3}+k)-(x_{3}+h)(y_{2}+k)+\) \((x_{3}+h)(y_{1}+k)-(x_{1}+h)(y_{3}+k)\}\)
\(=\frac{1}{2}\{x_{1}y_{2}+kx_{1}+hy_{2}+hk-x_{2}y_{1}-kx_{2}-hy_{1}-hk+x_{2}y_{3}+kx_{2}+\)
\(hy_{3}+hk-x_{3}y_{2}-kx_{3}-hy_{2}-hk+x_{3}y_{1}+kx_{3}+hy_{1}+hk-x_{1}y_{3}-kx_{1}-hy_{3}-hk\}\)
\(\therefore \triangle PQR=\frac{1}{2}\{x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{1}y_{3}\}\) বর্গ একক।
\(\therefore \triangle ABC= \triangle PQR\)
[দেখানো হলো।]

মনে করি, \(A(3, 0)\), \(B(0, 7)\), \(C(1, 1)\) এবং \(P(13, 3)\), \(Q(2, 3)\), \(R(-11, 2)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}3 \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ 3\\ 0 \ \ \ \ 7 \ \ \ \ 1 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}(3.7-0.0+0.1-1.7+1.0-3.1)\)
\(=\frac{1}{2}(21-0+0-7+0-3)\)
\(=\frac{1}{2}(21-10)\)
\(=\frac{1}{2}\times 11\)
\(=\frac{11}{2}\) বর্গ একক।
\(\therefore \triangle PQR=\frac{1}{2}\left|\begin{array}{c}13 \ \ \ \ 2 \ \ -11 \ \ \ \ 13\\ 3 \ \ \ \ \ \ 3 \ \ \ \ \ \ 2 \ \ \ \ \ \ 3\end{array}\right|\)
\(=\frac{1}{2}\{13.3-2.3+2.2-(-11).3+(-11).3-13.2\}\)
\(=\frac{1}{2}\{39-6+4+33-33-26\}\)
\(=\frac{1}{2}(76-65)\)
\(=\frac{1}{2}\times 11\)
\(=\frac{11}{2}\) বর্গ একক।
\(\therefore \triangle ABC=\triangle PQR\)
[দেখানো হলো।]

মনে করি, \(A(-1, 3)\), \(B(2, 9)\) এবং \(C(-3, -1)\)
\(\therefore \delta_{ABC}=\left|\begin{array}{c}-1 \ \ \ \ 2 \ \ -3 \ \ -1\\ 3 \ \ \ \ 9 \ \ -1 \ \ \ \ \ 3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(=(-1).9-2.3+2.(-1)-(-3).9+(-3).3-(-1)(-1)\)
\(=-9-6-2+27-9-1\)
\(=27-27\)
\(=0\)
\(\because \delta_{ABC}=0 \therefore \) প্রদত্ত বিন্দু তিনটি একই সরলরেখায় অবস্থিত।

মনে করি, \(A(k,-1)\), \(B(2, 3)\) এবং \(C(0, 1)\)
\(\therefore \delta_{ABC}=\left|\begin{array}{c}k \ \ \ \ 2 \ \ \ \ 0 \ \ \ \ k\\ -1 \ \ \ \ 3 \ \ \ \ 1 \ \ -1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(=k.3-2.(-1)+2.1-0.3+0.(-1)-k.1\)
\(=3k+2+2-0+0-k\)
\(=2k+4\)
\(\because \) প্রদত্ত বিন্দু তিনটি একই সরলরেখায় অবস্থিত।
\(\therefore \delta_{ABC}=0\)
\(\Rightarrow 2k+4=0\)
\(\Rightarrow 2k+4=0\)
\(\Rightarrow 2k=-4\)
\(\Rightarrow k=\frac{-4}{2}\)
\(\therefore k=-2\).

মনে করি, \(A(x, y)\), \(B(1, 2)\) এবং \(C(2, 1)\)
শর্তমতে,
\(\triangle ABC=6\)
\(\Rightarrow \frac{1}{2}\left|\begin{array}{c}x \ \ \ \ 1 \ \ \ \ 2 \ \ \ \ x\\ y \ \ \ \ 2 \ \ \ \ 1 \ \ \ \ y\end{array}\right|=6\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \left|\begin{array}{c}x \ \ \ \ 1 \ \ \ \ 2 \ \ \ \ x\\ y \ \ \ \ 2 \ \ \ \ 1 \ \ \ \ y\end{array}\right|=6\times 2\)
\(\Rightarrow x.2-1.y+1.1-2.2+2.y-x.1=12\)
\(\Rightarrow 2x-y+1-4+2y-x=12\)
\(\Rightarrow x+y-3-12=0\)
\(\Rightarrow x+y-15=0\)
\(\therefore x+y=15\)
[দেখানো হলো।]

দেওয়া আছে, \(A(-1, 2)\), \(B(2, 3)\), \(C(3, -4)\) এবং \(P(x, y)\)
এখন,
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}-1 \ \ \ \ 2 \ \ \ \ 3 \ \ -1\\ 2 \ \ \ \ 3 \ \ -4 \ \ \ \ 2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-1).3-2.2+2.(-4)-3.3+3.2-(-1)(-4)\}\)
\(=\frac{1}{2}\{-3-4-8-9+6-4\}\)
\(=\frac{1}{2}\{-3-4-8-9+6-4\}\)
\(=\frac{1}{2}\{6-28\}\)
\(=\frac{1}{2}\{-22\}\)
\(=-11\)
\(\therefore \triangle ABC=11\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)
আবার,
\(\triangle PAB=\frac{1}{2}\left|\begin{array}{c}x \ \ -1 \ \ \ \ 2 \ \ \ \ x\\ y \ \ \ \ 2 \ \ \ \ 3 \ \ \ \ y\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{x.2-(-1).y+(-1).3-2.2+2.y-x.3\}\)
\(=\frac{1}{2}\{2x+y-3-4+2y-3x\}\)
\(=\frac{1}{2}\{-x+3y-7\}\)
\(=\frac{1}{2}\{-(x-3y+7)\}\)
\(=-\frac{1}{2}(x-3y+7)\)
\(\therefore \triangle PAB=\frac{1}{2}(x-3y+7)\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)
এখন,
\(\frac{\triangle PAB}{\triangle ABC}=\frac{\frac{1}{2}(x-3y+7)}{11}\)
\(\Rightarrow \frac{\triangle PAB}{\triangle ABC}=\frac{(x-3y+7)}{2\times 11}\)
\(\therefore \frac{\triangle PAB}{\triangle ABC}=\frac{x-3y+7}{22}\)
[দেখানো হলো।]

মনে করি, \(A(a, b)\), \(B(b, a)\) এবং \(C(\frac{1}{a}, \frac{1}{b})\)
শর্তমতে,
\(\delta_{ABC}=0\)
\(\therefore \delta_{ABC}=\left|\begin{array}{c}a \ \ \ \ b \ \ \ \ \frac{1}{a} \ \ \ \ a\\ b \ \ \ \ a \ \ \ \ \frac{1}{b}\ \ \ \ b\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow a.a-b.b+b.\frac{1}{b}-\frac{1}{a}.a+\frac{1}{a}.b-a.\frac{1}{b}=0\)
\(\Rightarrow a^{2}-b^{2}+1-1+\frac{b}{a}-\frac{a}{b}=0\)
\(\Rightarrow a^{2}-b^{2}+\frac{b^{2}-a^{2}}{ab}=0\)
\(\Rightarrow a^{2}-b^{2}-\frac{a^{2}-b^{2}}{ab}=0\)
\(\Rightarrow (a^{2}-b^{2})(1-\frac{1}{ab})=0\)
\(\Rightarrow (a+b)(a-b)(1-\frac{1}{ab})=0\)
\(\Rightarrow a+b=0, a-b\neq 0, 1-\frac{1}{ab}\neq 0\) ➜Note \(\because a-b=0, 1-\frac{1}{ab}=0\) হলে, বিন্দুগুলির ভিন্নতা থাকে না।
\(\therefore a+b=0\)
[দেখানো হলো।]

মনে করি, \(A(a, 0)\), \(B(0, b)\) এবং \(C(1, 1)\)
শর্তমতে,
\(\delta_{ABC}=0\)
\(\therefore \delta_{ABC}=\left|\begin{array}{c}a \ \ \ \ 0 \ \ \ \ 1 \ \ \ \ a\\ 0 \ \ \ \ b \ \ \ \ 1\ \ \ \ 0\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(\Rightarrow a.b-0.0+0.1-1.b+1.0-a.1=0\)
\(\Rightarrow ab-0+0-b+0-a=0\)
\(\Rightarrow ab-b-a=0\)
\(\Rightarrow ab=b+a\)
\(\Rightarrow \frac{ab}{ab}=\frac{b+a}{ab}\) ➜Note উভয় পার্শে \(ab\) ভাগ করে।
\(\Rightarrow 1=\frac{b}{ab}+\frac{a}{ab}\)
\(\Rightarrow 1=\frac{1}{a}+\frac{1}{b}\)
\(\therefore \frac{1}{a}+\frac{1}{b}=1\)
[দেখানো হলো।]

মনে করি, \(A(x, y)\), \(B(5, 3)\) এবং \(C(-2, -4)\)
শর্তমতে,
\(\delta_{ABC}=0\)
\(\therefore \delta_{ABC}=\left|\begin{array}{c}x \ \ \ \ 5 \ \ -2 \ \ \ \ x\\ y \ \ \ \ 3 \ \ -4 \ \ \ \ y\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(\Rightarrow x.3-5.y+5.(-4)-(-2).3+(-2).y-x.(-4)=0\)
\(\Rightarrow 3x-5y-20+6-2y+4x=0\)
\(\Rightarrow 7x-7y-14=0\)
\(\therefore x-y-2=0\) ➜Note উভয় পার্শে \(7\) ভাগ করে।
[ দেখানো হলো। ]

মনে করি, \(A(a^{2}, bc)\), \(B(b^{2}, ca)\) এবং \(C(c^{2}, ab)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}a^{2} \ \ \ \ b^{2} \ \ \ \ c^{2} \ \ \ \ a^{2}\\ bc \ \ \ \ ca\ \ \ \ ab\ \ \ \ bc\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{a^{2}.ca-b^{2}.bc+b^{2}.ab-c^{2}.ca+c^{2}.bc-a^{2}.ab\}\)
\(=\frac{1}{2}\{a^{3}c-b^{3}c+ab^{3}-c^{3}a+bc^{3}-a^{3}b\}\)
\(=\frac{1}{2}\{c(a^{3}-b^{3})-ab(a^{2}-b^{2})-c^{3}(a-b)\}\)
\(=\frac{1}{2}(a-b)\{c(a^{2}+ab+b^{2})-ab(a+b)-c^{3}\}\)
\(=\frac{1}{2}(a-b)\{ca^{2}+abc+b^{2}c-a^{2}b-ab^{2}-c^{3}\}\)
\(=\frac{1}{2}(a-b)\{b^{2}(c-a)-c(c^{2}-a^{2})+ab(c-a)\}\)
\(=\frac{1}{2}(a-b)(c-a)\{b^{2}-c(c+a)+ab\}\)
\(=\frac{1}{2}(a-b)(c-a)\{b^{2}-c^{2}-ca+ab\}\)
\(=\frac{1}{2}(a-b)(c-a)\{(b^{2}-c^{2})+a(b-c)\}\)
\(=\frac{1}{2}(a-b)(b-c)(c-a)\{b+c+a\}\)
\(=\frac{1}{2}(a-b)(b-c)(c-a)(a+b+c)\)
\(\therefore \triangle ABC=\frac{1}{2}(a-b)(b-c)(c-a)(a+b+c)\) বর্গ একক।
যদি,
\(a+b+c=0\) হয়।
\(\therefore \triangle ABC=\frac{1}{2}(a-b)(b-c)(c-a)\times 0\)
\(\Rightarrow \triangle ABC=0\)
\(\because \delta_{ABC}=2\triangle ABC\)
\(\Rightarrow \delta_{ABC}=2\times 0\)
\(\Rightarrow \delta_{ABC}=0\)
\(\therefore A, B, C\) বিন্দুগুলি সমরেখ।
[ দেখানো হলো। ]

মনে করি, \(A(p, p-2)\), \(B(p+3, p)\) এবং \(C(p+2, p+2)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}p \ \ \ \ p+3 \ \ \ \ p+2 \ \ \ \ p\\p-2 \ \ \ \ p \ \ \ \ p+2\ \ p-2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{p.p-(p+3)(p-2)+(p+3)(p+2)-(p+2).p+(p+2)(p-2)-p.(p+2)\}\)
\(=\frac{1}{2}\{p^{2}-p^{2}-3p+2p+6+p^{2}+3p+2p+6-p^{2}-2p+p^{2}-4-p^{2}-2p\}\)
\(=\frac{1}{2}\times 8\)
\(=4\) বর্গ একক। যাহা \(p\) বর্জিত।
[ প্রমাণিত ]

দেওয়া আছে, \(A(5, 6)\), \(B(-9, 1)\) এবং \(C(-3, -1)\)।
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}5 \ \ -9 \ \ -3 \ \ \ \ \ 5\\6 \ \ \ \ \ 1 \ \ -1 \ \ \ \ \ 6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{5.1-(-9).6+(-9).(-1)-(-3).1+(-3).6-5.(-1)]\)
\(=\frac{1}{2}\{5+54+9+3-18+5]\)
\(=\frac{1}{2}\{76-18\}\)
\(=\frac{1}{2}\times 58\)
\(=29\)
\(\therefore \triangle ABC=29\) বর্গ একক।
এখন,
\(BC=\sqrt{(-9+3)^{2}+(1+1)^{2}}\)
\(=\sqrt{6^{2}+2^{2}}\)
\(CA=\sqrt{36+4}\)
\(CA=\sqrt{10\times 4}\)
\(CA=\sqrt{10\times 2^{2}}\)
\(\therefore CA=2\sqrt{10}\)
\(A\) বিন্দু হতে \(BC\) বাহুর উপর অঙ্কিত লম্বের দৈর্ঘ্য \(AD\)
\(\therefore \frac{1}{2}\times BC\times AD=\triangle ABC\)
\(\Rightarrow \frac{1}{2}\times 2\sqrt{10}\times AD=29\)
\(\Rightarrow \sqrt{10} AD=29\)
\(\Rightarrow AD=\frac{29}{\sqrt{10}}\)
\(\therefore AD=\frac{29\times \sqrt{10}}{(\sqrt{10})^{2}}\)
\(\therefore AD=\frac{29}{10}\sqrt{10}\) একক।

দেওয়া আছে, \(A(6, 3)\), \(B(-3, 5)\) এবং \(C(4, 2)\)
\(D\), \(BC\) কে \(3:1\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore D(\frac{3.4+1.(-3)}{3+1}, \frac{3.2+1.5}{3+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow D(\frac{12-3}{4}, \frac{6+5}{4})\)
\(\therefore D(\frac{9}{4}, \frac{11}{4})\)
\(E\), \(CA\) কে \(3:1\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore E(\frac{3.6+1.4}{3+1}, \frac{3.3+1.2}{3+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow E(\frac{18+4}{4}, \frac{9+2}{4})\)
\(\Rightarrow E(\frac{22}{4}, \frac{11}{4})\)
\(\therefore E(\frac{11}{2}, \frac{11}{4})\)
\(F\), \(AB\) কে \(3:1\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore F(\frac{3.(-3)+1.6}{3+1}, \frac{3.5+1.3}{3+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow F(\frac{-9+6}{4}, \frac{15+3}{4})\)
\(\Rightarrow F(\frac{-3}{4}, \frac{18}{4})\)
\(\therefore F(-\frac{3}{4}, \frac{9}{2})\)
এখন,
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}6 \ \ -3 \ \ \ \ 4 \ \ \ \ 6\\ 3 \ \ \ \ 5\ \ \ \ 2\ \ \ \ 3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{6.5-(-3).3+(-3).2-4.5+4.3-6.2\}\)
\(=\frac{1}{2}\{30+9-6-20+12-12\}\)
\(=\frac{1}{2}\{39-26\}\)
\(=\frac{1}{2}\times 13\)
\(\therefore \triangle ABC=\frac{13}{2}\) বর্গ একক।
আবার,
\(\therefore \triangle DEF=\frac{1}{2}\left|\begin{array}{c}\frac{9}{4} \ \ \ \ \ \frac{11}{2} \ \ -\frac{3}{4} \ \ \ \ \frac{9}{4}\\\frac{11}{4} \ \ \ \ \frac{11}{4} \ \ \ \ \frac{9}{2} \ \ \ \ \frac{11}{4}\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{\frac{9}{4}\times \frac{11}{4}-\frac{11}{2}\times \frac{11}{4}+\frac{11}{2}\times \frac{9}{2}-(-\frac{3}{4})\times \frac{11}{4}+(-\frac{3}{4})\times \frac{11}{4}-\frac{9}{4}\times \frac{9}{2}\}\)
\(=\frac{1}{2}\{\frac{99}{16}-\frac{121}{8}+\frac{99}{4}+\frac{33}{16}-\frac{33}{16}-\frac{81}{8}\}\)
\(=\frac{1}{2}\times \frac{99-121\times 2+99\times 4+33-33-81\times 2}{16}\)
\(=\frac{1}{2}\times \frac{99-242+396+33-33-162}{16}\)
\(=\frac{1}{2}\times \frac{495-404}{16}\)
\(=\frac{1}{2}\times \frac{91}{16}\)
\(=\frac{91}{32}\)
\(\therefore \triangle DEF=\frac{91}{32}\) বর্গ একক।
এখন,
\(\frac{\triangle ABC}{\triangle DEF}=\frac{\frac{13}{2}}{\frac{91}{32}}\)
\(=\frac{13}{2}\times \frac{32}{91}\)
\(=\frac{16}{7}\)
\(\Rightarrow \frac{\triangle ABC}{\triangle DEF}=\frac{16}{7}\)
\(\therefore \triangle ABC:\triangle DEF=16:7\)

দেওয়া আছে, \(A(3, 5)\), \(B(-3, 3)\) এবং \(C(-1, -1)\)
\(D\), \(BC\) এর মধ্যবিন্দু \(\therefore D(\frac{-3-1}{2}, \frac{3-1}{2})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\)-এর মধ্যবিন্দু \(\therefore R\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
\(\Rightarrow D(\frac{-4}{2}, \frac{2}{2})\)
\(\therefore D(-2, 1)\)
\(E\), \(CA\) এর মধ্যবিন্দু \(\therefore E(\frac{-1+3}{2}, \frac{-1+5}{2})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\)-এর মধ্যবিন্দু \(\therefore R\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
\(\Rightarrow E(\frac{2}{2}, \frac{4}{2})\)
\(\therefore E(1, 2)\)
\(F\), \(AB\) এর মধ্যবিন্দু \(\therefore F(\frac{3-3}{2}, \frac{5+3}{2})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\)-এর মধ্যবিন্দু \(\therefore R\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
\(\Rightarrow F(\frac{0}{2}, \frac{8}{2})\)
\(\therefore F(0, 4)\)
এখন,
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}3 \ \ -3 \ \ \ \ -1 \ \ \ \ 3\\ 5 \ \ \ \ \ 3 \ \ -1\ \ \ \ \ 5\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{3.3-(-3).5+(-3)(-1)-(-1).3+(-1).5-3.(-1)\}\)
\(=\frac{1}{2}\{9+15+3+3-5+3\}\)
\(=\frac{1}{2}\{33-5\}\)
\(=\frac{1}{2}\times 28\)
\(=14\)
\(\therefore \triangle ABC=14\) বর্গ একক।
আবার,
\(\therefore \triangle DEF=\frac{1}{2}\left|\begin{array}{c}-2 \ \ \ \ 1 \ \ \ \ 0 \ \ -2\\1 \ \ \ \ \ 2 \ \ \ \ 4 \ \ \ \ \ 1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-2).2-1.1+1.4-0.2+0.1-(-2).4\}\)
\(=\frac{1}{2}\{-4-1+4-0+0+8\}\)
\(=\frac{1}{2}\{12-5\}\)
\(=\frac{1}{2}\times 7\)
\(=\frac{7}{2}\)
\(=3.5\)
\(\therefore \triangle DEF=3.5\) বর্গ একক।
এখন,
\(\triangle ABC=14\)
\(\Rightarrow \triangle ABC=4\times 3.5\)
\(\therefore \triangle ABC=4\triangle DEF\)
[ দেখানো হলো। ]

দেওয়া আছে, \(A(4, -3)\), \(B(13, 0)\), \(C(-2, 9)\)
এবং \(\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}=2\)
\(\Rightarrow BD:DC=2:1, CE:EA=2:1, AF:FB=2:1\)
অর্থাৎ \(D\), \(BC\) কে \(2:1\) অনুপাতে অন্তর্বিভক্ত করে \(D(\frac{2.(-2)+1.13}{2+1}, \frac{2.9+1.0}{2+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow D(\frac{-4+13}{3}, \frac{18+0}{3})\)
\(\Rightarrow D(\frac{9}{3}, \frac{18}{3})\)
\(\therefore D(3, 6)\)
\(E\), \(CA\) কে \(2:1\) অনুপাতে অন্তর্বিভক্ত করে \(E(\frac{2.4+1.(-2)}{2+1}, \frac{2.(-3)+1.9}{2+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow E(\frac{8-2}{3}, \frac{-6+9}{3})\)
\(\Rightarrow E(\frac{6}{3}, \frac{3}{3})\)
\(\therefore E(2, 1)\)
\(F\), \(AB\) কে \(2:1\) অনুপাতে অন্তর্বিভক্ত করে \(F(\frac{2.13+1.4}{2+1}, \frac{2.0+1.(-3)}{2+1})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে অন্তর্বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow F(\frac{26+4}{3}, \frac{0-3}{3})\)
\(\Rightarrow F(\frac{30}{3}, \frac{-3}{3})\)
\(\therefore F(10, -1)\)
এখন,
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}4 \ \ \ \ 13 \ \ -2 \ \ \ \ 4\\ -3 \ \ \ \ \ 0 \ \ \ \ 9\ \ \ -3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{4.0-13.(-3)+13.9-(-2).0+(-2).(-3)-4.9\}\)
\(=\frac{1}{2}\{0+39+117+0+6-36\}\)
\(=\frac{1}{2}\{162-36\}\)
\(=\frac{1}{2}\times 126\)
\(=63\)
\(\therefore \triangle ABC=63\) বর্গ একক।
আবার,
\(\therefore \triangle DEF=\frac{1}{2}\left|\begin{array}{c}3 \ \ \ \ 2 \ \ \ \ 10 \ \ \ \ 3\\6 \ \ \ \ \ 1 \ \ -1 \ \ \ \ \ 6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{3.1-2.6+2.(-1)-10.1+10.6-3.(-1)\}\)
\(=\frac{1}{2}\{3-12-2-10+60+3\}\)
\(=\frac{1}{2}\{66-24\}\)
\(=\frac{1}{2}\times 42\)
\(=21\)
\(\therefore \triangle DEF=21\) বর্গ একক।
এখন,
\(\frac{\triangle ABC}{\triangle DEF}=\frac{63}{21}\)
\(\Rightarrow \frac{\triangle ABC}{\triangle DEF}=\frac{3}{1}\)
\(\therefore \triangle ABC:\triangle DEF=3:1\)
[ প্রমাণিত। ]

দেওয়া আছে, \(A(2, 5)\), \(B(-3, -1)\) এবং \(C(11, 9)\);
\(D\), \(BC\) এর মধ্যবিন্দু \(\therefore D(\frac{-3+11}{2}, \frac{-1+9}{2})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\)-এর মধ্যবিন্দু \(\therefore R\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
\(\Rightarrow D(\frac{8}{2}, \frac{8}{2})\)
\(\therefore D(4, 4)\)
\(AD\) মধ্যমার দৈর্ঘ্য \(=\sqrt{(2-4)^{2}+(5-4)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-2)^{2}+1^{2}}\)
\(=\sqrt{4+1}\)
\(=\sqrt{5}\) একক।
এখন,
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}2 \ \ -3 \ \ \ \ 11 \ \ \ \ 2\\5 \ \ \ -1 \ \ \ \ 9 \ \ \ \ 5\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{2.(-1)-(-3).5+(-3).9-11.(-1)+11.5-2.9\}\)
\(=\frac{1}{2}\{-2+15-27+11+55-18\}\)
\(=\frac{1}{2}\{81-47\}\)
\(=\frac{1}{2}\times 34\)
\(=17\)
\(\therefore \triangle ABC=17\) বর্গ একক।
আবার,
\(\therefore \triangle ABD=\frac{1}{2}\left|\begin{array}{c}2 \ \ -3 \ \ \ \ 4 \ \ \ \ 2\\5 \ \ \ -1 \ \ \ \ 4 \ \ \ \ 5\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{2.(-1)-(-3).5+(-3).4-4.(-1)+4.5-2.4\}\)
\(=\frac{1}{2}\{-2+15-12+4+20-8\}\)
\(=\frac{1}{2}\{39-22\}\)
\(=\frac{1}{2}\times 17\)
\(=\frac{17}{2}\)
\(\therefore \triangle ABD=\frac{17}{2}\) বর্গ একক।
এখন,
\(\triangle ABD=\frac{17}{2}\)
\(\Rightarrow \triangle ABD=\frac{1}{2}\times 17\)
\(\therefore \triangle ABD=\frac{1}{2}\triangle ABC\)
[ দেখানো হলো। ]

দেওয়া আছে, \(O(0, 0)\), \(P(A\cos\beta, -A\sin\beta)\) এবং \(Q(A\sin\alpha, A\cos\alpha)\)
\(\therefore \triangle OPQ=\frac{1}{2}\left|\begin{array}{c}0 \ \ A\cos\beta \ \ A\sin\alpha \ \ \ \ 0\\0 \ \ -A\sin\beta \ \ A\cos\alpha\ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.(-A\sin\beta)-A\cos\beta.0+A\cos\beta.A\cos\alpha-A\sin\alpha.(-A\sin\beta)+A\sin\alpha.0-0.A\cos\alpha]\)
\(=\frac{1}{2}\{0-0+A^{2}\cos\alpha\cos\beta+A^{2}\sin\alpha\sin\beta+0-0\}\)
\(=\frac{1}{2}\{A^{2}\cos\alpha\cos\beta+A^{2}\sin\alpha\sin\beta\}\)
\(=\frac{1}{2}A^{2}(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\)
\(=\frac{1}{2}A^{2}\cos(\alpha-\beta)\)
\(\therefore \triangle OPQ=\frac{1}{2}A^{2}\cos(\alpha-\beta)\)
এখানে, \(\frac{1}{2}A^{2}\) ধ্রুবক। \(\therefore \triangle OPQ\) বৃহত্তম হবে যদি \(\cos(\alpha-\beta)\) বৃহত্তম হয়।
\(\cos(\alpha-\beta)\) এর বৃহত্তম মান \(1\) ➜Note \(\because 1 > \cos\theta > -1\)
\(\therefore \cos(\alpha-\beta)=1\) হলে \(\triangle OPQ\) বৃহত্তম হবে।
\(\Rightarrow \cos(\alpha-\beta)=0\)
\(\Rightarrow \cos(\alpha-\beta)=\cos0\)
\(\Rightarrow \alpha-\beta=0\)
\(\Rightarrow \alpha=\beta\)
\(\therefore \alpha=\beta\) হলে, ত্রিভুজটির ক্ষেত্রফলের মান বৃহত্তম হবে।
ক্ষেত্রফলের বৃহত্তম মান \(=\frac{1}{2}A^{2}.1\)
\(=\frac{1}{2}A^{2}\) বর্গ একক।

মনে করি, \(A(t+1, 1)\), \(B(2t+1, 3)\) এবং \(C(2t+2, 2t)\)
\(\therefore \triangle ABC=\frac{1}{2}\left|\begin{array}{c}t+1 \ \ 2t+1 \ \ 2t+2 \ \ t+1\\1 \ \ \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ 2t \ \ \ \ \ \ \ \ \ \ 1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(t+1).3-(2t+1).1+(2t+1).2t-(2t+2).3+(2t+2).1-(t+1).2t\}\)
\(=\frac{1}{2}(3t+3-2t-1+4t^{2}+2t-6t-6+2t+2-2t^{2}-2t)\)
\(=\frac{1}{2}(2t^{2}-3t-2)\)
\(\therefore \triangle ABC=\frac{1}{2}(2t^{2}-3t-2)\)
এখন, বিন্দুগুলি সমরেখ হবে যদি \(\triangle ABC=0\) হয়।
\(\therefore \triangle ABC=0\)
\(\Rightarrow \frac{1}{2}(2t^{2}-3t-2)=0\)
\(\Rightarrow 2t^{2}-3t-2=0\)
\(\Rightarrow 2t^{2}-4t+t-2=0\)
\(\Rightarrow 2t(t-2)+1(t-2)=0\)
\(\Rightarrow (t-2)(2t+1)=0\)
\(\Rightarrow t-2=0, 2t+1=0\)
\(\Rightarrow t=2, 2t=-1\)
\(\therefore t=2, t=-\frac{1}{2}\)
\(\therefore t=2\) বা \(t=-\frac{1}{2}\) হলে বিন্দুগুলি সমরেখ হবে।
[ দেখানো হলো। ]

মনে করি, \(A(3, 90^{o})\), \(B(3, 30^{o})\) এবং মূলবিন্দু \(O(0, 0)\)
\(A(3, 90^{o})\) এর কার্তেসীয় স্থানাঙ্ক \(A(3\cos90^{o}, 3\sin90^{o})\)
\(\Rightarrow A(3\times 0, 3\times 1)\)
\(\therefore A(0, 3)\)
\(B(3, 30^{o})\) এর কার্তেসীয় স্থানাঙ্ক \(B(3\cos30^{o}, 3\sin30^{o})\)
\(\Rightarrow B(3\times \frac{\sqrt{3}}{2}, 3\times \frac{1}{2})\)
\(\therefore B(\frac{3\sqrt{3}}{2}, \frac{3}{2})\)
এখন,
\(AO=\sqrt{(0-0)^{2}+(3-0)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{0^{2}+3^{2}}\)
\(=\sqrt{0+9}\)
\(=\sqrt{9}\)
\(=\sqrt{3^{2}}\)
\(=3\)
\(BO=\sqrt{(\frac{3\sqrt{3}}{2}-0)^{2}+(\frac{3}{2}-0)^{2}}\)
\(=\sqrt{(\frac{3\sqrt{3}}{2})^{2}+(\frac{3}{2})^{2}}\)
\(=\sqrt{\frac{27}{4}+\frac{9}{4}}\)
\(=\sqrt{\frac{27+9}{4}}\)
\(=\sqrt{\frac{36}{4}}\)
\(=\sqrt{9}\)
\(=3\)
\(AB=\sqrt{(0-\frac{3\sqrt{3}}{2})^{2}+(3-\frac{3}{2})^{2}}\)
\(=\sqrt{\frac{(3\sqrt{3}}{2})^{2}+(\frac{6-3}{2})^{2}}\) \(=\sqrt{\frac{27}{4}+(\frac{3}{2})^{2}}\)
\(=\sqrt{\frac{27}{4}+\frac{9}{4}}\)
\(=\sqrt{\frac{27+9}{4}}\)
\(=\sqrt{\frac{36}{4}}\)
\(=\sqrt{9}\)
\(=3\)
\(\because AO=BO=AB \therefore \triangle OAB\) সমবাহু।
আবার,
\(\triangle OAB=\frac{1}{2}\left|\begin{array}{c}0 \ \ \ \ \ \ 0 \ \ \ \ \frac{3\sqrt{3}}{2} \ \ \ \ \ \ \ 0\\0 \ \ \ \ \ \ 3 \ \ \ \ \ \ \ \ \frac{3}{2} \ \ \ \ \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.3-0.0+0.\frac{3}{2}-\frac{3\sqrt{3}}{2}.3+\frac{3\sqrt{3}}{2}.0-0.\frac{3}{2}]\)
\(=\frac{1}{2}\{0-0+0-\frac{9\sqrt{3}}{2}+0-0\}\)
\(=\frac{1}{2}(-\frac{9\sqrt{3}}{2})\)
\(=-\frac{9}{4}\sqrt{3}\)
\(\therefore \triangle OAB=\frac{9}{4}\sqrt{3}\) ➜Note \(\because \triangle \neq -ve\)

দেওয়া আছে, \(A(2, 6)\),\(B(-7, -3)\) এবং \(C(5,-6)\)
\(\therefore \triangle ABC\) এর ভরকেন্দ্র \(G(\frac{2-7+5}{3}, \frac{6-3-6}{3})\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, এর ভরকেন্দ্র \(G(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3})\) হয়।
\(\Rightarrow G(\frac{0}{3}, \frac{-3}{3})\)
\(\Rightarrow G(0, -1)\)
\(\therefore \triangle ABC\) এর ভরকেন্দ্র \(G(0, -1)\)
এখন,
\(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}2 \ \ -7 \ \ \ \ \ 5 \ \ \ \ \ 2\\6 \ \ -3 \ \ -6 \ \ \ \ 6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{2.(-3)-(-7).6+(-7).(-6)-5.(-3)+5.6-2.(-6)\}\)
\(=\frac{1}{2}(-6+42+42+15+30+12)\)
\(=\frac{1}{2}(141-6)\)
\(=\frac{1}{2}\times 135\)
\(\therefore \triangle ABC=\frac{135}{2}\) বর্গ একক।
\(\triangle ABG=\frac{1}{2}\left|\begin{array}{c}2 \ \ -7 \ \ \ \ \ 0 \ \ \ \ \ 2\\6 \ \ -3 \ \ -1 \ \ \ \ 6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{2.(-3)-(-7).6+(-7).(-1)-0.(-3)+0.6-2.(-1)\}\)
\(=\frac{1}{2}(-6+42+7+0+0+2)\)
\(=\frac{1}{2}(51-6)\)
\(=\frac{1}{2}\times 45\)
\(\therefore \triangle ABG=\frac{45}{2}\) বর্গ একক।
\(\triangle BCG=\frac{1}{2}\left|\begin{array}{c}-7 \ \ \ \ \ 5 \ \ \ \ \ 0 \ \ -7\\-3 \ \ -6 \ \ -1 \ \ -3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{(-7).(-6)-5.(-3)+5.(-1)-0.(-6)+0.(-3)-(-7).(-1)\}\)
\(=\frac{1}{2}(42+15-5+0-0-7)\)
\(=\frac{1}{2}(57-12)\)
\(=\frac{1}{2}\times 45\)
\(\therefore \triangle BCG=\frac{45}{2}\) বর্গ একক।
\(\triangle CAG=\frac{1}{2}\left|\begin{array}{c}5 \ \ \ \ \ 2 \ \ \ \ \ \ 0 \ \ \ \ \ \ 5\\-6 \ \ \ \ 6 \ \ -1 \ \ -6\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{5.6-2.(-6)+2.(-1)-0.6+0.(-6)-5.(-1)\}\)
\(=\frac{1}{2}(30+12-2+0-0+5)\)
\(=\frac{1}{2}(47-2)\)
\(=\frac{1}{2}\times 45\)
\(\therefore \triangle CAG=\frac{45}{2}\) বর্গ একক।
এখন,
\(\triangle ABC=\frac{135}{2}\)
\(\Rightarrow \triangle ABC=3\times \frac{45}{2}\)
\(\Rightarrow \triangle ABC=3\triangle ABG\)
\(\Rightarrow \triangle ABC=3\triangle BCG\)
\(\Rightarrow \triangle ABC=3\triangle CAG\)
\(\therefore \triangle ABC=3\triangle ABG=3\triangle BCG=3\triangle CAG\)
[ দেখানো হলো। ]

দেওয়া আছে, \(A(3, 5)\), \(B(3, 8)\) এবং মূলবিন্দু \(O(0, 0)\)
\(AB=\sqrt{(3-3)^{2}+(5-8)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{0^{2}+(-3)^{2}}\)
\(=\sqrt{0+9}\)
\(=\sqrt{9}\)
\(=3\)
\(AO=\sqrt{(3-0)^{2}+(5-0)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{3^{2}+5^{2}}\)
\(=\sqrt{9+25}\)
\(=\sqrt{34}\)
\(=5.83\)
\(BO=\sqrt{(3-0)^{2}+(8-0)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{3^{2}+8^{2}}\)
\(=\sqrt{9+64}\)
\(=\sqrt{73}\)
\(=8.54\)
এখন,
\(AB+AO=3+5.83=8.83>8.54\)
\(\Rightarrow AB+AO>8.54\)
\(\Rightarrow AB+AO>BO\)
\(\therefore OAB\) একটি ত্রিভুজ।
[ দেখানো হলো। ] \(\triangle OAB=\frac{1}{2}\left|\begin{array}{c}0 \ \ \ \ 3 \ \ \ \ 3 \ \ \ \ 0\\0 \ \ \ \ 5 \ \ \ \ 8 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.5-3.0+3.8-3.5+3.0-0.8\}\)
\(=\frac{1}{2}(0-0+24-15+0-0)\)
\(=\frac{1}{2}(24-15)\)
\(=\frac{1}{2}\times 9\)
\(=\frac{9}{2}\)
\(=4\frac{1}{2}\) বর্গ একক।

মনে করি, \(A(1, 2)\), \(B(-5, 6)\), \(C(7, -4)\) এবং \(D(k, -2)\)
শর্তমতে,\(\Box ABCD=0\)
\(\therefore \frac{1}{2}\left|\begin{array}{c}1 \ \ -5 \ \ \ \ \ 7 \ \ \ \ \ k \ \ \ \ \ 1\\2 \ \ \ \ 6 \ \ -4 \ \ -2 \ \ \ \ 2\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দু চারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \frac{1}{2}\{1.6-(-5).2+(-5).(-4)-7.6+7.(-2)-k.(-4)+k.2-1.(-2)\}=0\)
\(\Rightarrow (6+10+20-42-14+4k+2k+2)=0\times 2\)
\(\Rightarrow 6k+38-56=0\)
\(\Rightarrow 6k-18=0\)
\(\Rightarrow 6k=18\)
\(\Rightarrow k=\frac{18}{6}\)
\(\therefore k=3\)

দেওয়া আছে, \(A(0, -1)\), \(B(15, 2)\), \(C(-1, 2)\) এবং \(D(4, -5)\)।
মনে করি, \(AB\) সরলরেখা \(CD\) সরলরেখাকে \(P(x, y)\) বিন্দুতে \(m:n\) অনুপাতে বিভক্ত করে।
\(\therefore P(\frac{m\times 4+n\times -1}{m+n}, \frac{m\times -5+n\times 2}{m+n})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\) কে \(m:n\) অনুপাতে বিভক্ত করে। \(\therefore R\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)\)
\(\Rightarrow P(\frac{4m-n}{m+n}, \frac{-5m+2n}{m+n})\)
এখন,
\(A, B, P\) একই সরলরেখায় অবস্থিত।
\(\therefore \delta_{ABP}=0\)
\(\Rightarrow \left|\begin{array}{c}0 \ \ 15 \ \ \frac{4m-n}{m+n} \ \ \ \ 0\\ -1 \ \ 2 \ \ \frac{-5m+2n}{m+n} \ \ \ \ -1\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি একই সরলরেখায় অবস্থিত হলে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{1}\\ y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|=0\) হয়।
\(\Rightarrow 0.(2)-15.(-1)+15.(\frac{-5m+2n}{m+n})-2.(\frac{4m-n}{m+n})+\frac{4m-n}{m+n}.(-1)-0.(\frac{-5m+n}{m+n})=0\)
\(\Rightarrow 0+15+\frac{-75m+30n}{m+n}-\frac{8m-2n}{m+n}-\frac{4m-n}{m+n}+0=0\)
\(\Rightarrow 15m+15n-75m+30n-8m+2n-4m+n=0\) ➜Note উভয়পার্শে \((m+n)\) গুণ করে।
\(\Rightarrow -72m+48n=0\)
\(\Rightarrow -72m=-48n\)
\(\Rightarrow 3m=2n\) ➜Note উভয়পার্শে \((-24)\) ভাগ করে।
\(\Rightarrow \frac{m}{n}=\frac{2}{3}\)
\(\Rightarrow m:n=2:3\)
\(\therefore AB\) সরলরেখা \(CD\) সরলরেখাকে \(2:3\) অনুপাতে অন্তর্বিভক্ত করে।

দেওয়া আছে, \(A(t-4, -2)\), \(B(t, t+3)\), \(C(2t+1, 1)\), \(D(t-3, 1)\) এবং মূলবিন্দু \(O(0, 0)\)
\(\therefore \triangle OAB=\frac{1}{2}\left|\begin{array}{c}0 \ \ \ t-4 \ \ \ \ t \ \ \ \ \ 0\\0 \ \ -2 \ \ t+3 \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.(-2)-(t-4).0+(t-4).(t+3)-t.(-2)+t.0-0.(t+3)\}\)
\(=\frac{1}{2}(0-0+t^{2}-4t+3t-12+2t+0-0)\)
\(=\frac{1}{2}(t^{2}+t-12)\)
\(\therefore \triangle OAB=\frac{1}{2}(t^{2}+t-12)\) বর্গ একক।
\(\therefore \triangle OCD=\frac{1}{2}\left|\begin{array}{c}0 \ \ 2t+1 \ \ t-3 \ \ \ \ 0\\0 \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ 1 \ \ \ \ \ \ \ 0\end{array}\right|=0\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.1-(2t+1).0+(2t+1).1-(t-3).1+(t-3).0-0.1\}\)
\(=\frac{1}{2}(0-0+2t+1-t+3+0-0)\)
\(=\frac{1}{2}(t+4)\)
\(\therefore \triangle OCD=\frac{1}{2}(t+4)\) বর্গ একক।
এখন,
\(\frac{\triangle OAB}{\triangle OCD}=\frac{\frac{1}{2}(t^{2}+t-12)}{\frac{1}{2}(t+4)}\)
\(=\frac{t^{2}+t-12}{t+4}\)
\(=\frac{t^{2}+4t-3t-12}{t+4}\)
\(=\frac{t(t+4)-3(t+4)}{t+4}\)
\(=\frac{(t+4)(t-3)}{t+4}\)
\(=t-3\)
\(\therefore \frac{\triangle OAB}{\triangle OCD}=(t-3)\)
\(\Rightarrow \triangle OAB:\triangle OCD=(t-3):1\)
[ দেখানো হলো। ]
আবার,
\(\triangle OAB=\frac{1}{2}(t^{2}+t-12)\)
\(\Rightarrow \triangle OAB=\frac{1}{2}(4^{2}+4-12)\) ➜Note \(\because t=4\)
\(\Rightarrow \triangle OAB=\frac{1}{2}(16+4-12)\)
\(\Rightarrow \triangle OAB=\frac{1}{2}(20-12)\)
\(\Rightarrow \triangle OAB=\frac{1}{2}\times 8\)
\(\Rightarrow \triangle OAB=4\)
\(\triangle OCD=\frac{1}{2}(t+4)\)
\(\Rightarrow \triangle OCD=\frac{1}{2}(4+4)\) ➜Note \(\because t=4\)
\(\Rightarrow \triangle OCD=\frac{1}{2}\times 8\)
\(\Rightarrow \triangle OCD=4\)
\(\therefore t=4\) হলে, ত্রিভুজ দ্বয়ের ক্ষেত্রফলের মান সমান এবং সমচিহ্ন বিশিষ্ট হবে ।

মনে করি, \(A(-2, 3)\), \(B(-3, -4)\), \(C(5, -1)\) এবং \(D(2, 2)\)
\(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}-2 \ \ -3 \ \ \ \ \ 5 \ \ \ \ \ 2 \ \ -2\\3 \ \ -4 \ \ -1 \ \ \ \ 2 \ \ \ \ 3\end{array}\right|\)
\(=\frac{1}{2}\{(-2).(-4)-(-3).3+(-3).(-1)-5.(-4)+5.2-2.(-1)+2.3-(-2).2\}\)
\(=\frac{1}{2}(8+9+3+20+10+2+6+4)\)
\(=\frac{1}{2}\times 62\)
\(=31\) বর্গ একক।

মনে করি, \(A(a, 0)\), \(B(-b, 0)\), \(C(0, a)\) এবং \(D(0, -b)\)
\(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}a \ \ -b \ \ \ \ \ 0 \ \ \ \ \ 0 \ \ \ \ a\\0 \ \ \ \ 0 \ \ \ \ a \ \ -b \ \ \ \ 0\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দু চারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{a.0-(-b).0+(-b).a-0.0+0.(-b)-0.a+0.0-a.(-b)\}\)
\(=\frac{1}{2}(0+0-ab-0-0-0+0+ab)\)
\(=\frac{1}{2}(-ab+ab)\)
\(=\frac{1}{2}\times 0\)
\(=0\)
\(\therefore \Box ABCD=0\)
[ দেখানো হলো। ]
ব্যাখ্যাঃ বিন্দুগুলি সমরেখ তাই ,বিন্দু চারটি দ্বারা গঠিত চতুর্ভজের ক্ষেত্রফল শুন্য।

দেওয়া আছে, \(A(1, 2)\), \(B(3, 4)\), \(C(1, 0)\).
ধরি, \(D(x, y)\)
\((a)\) \(AC\) এর মধ্যবিন্দু \(E(\frac{1+1}{2}, \frac{2+0}{2})\) ➜Note \(P(x_1, y_1)\), \(Q(x_2, y_2)\); R, \(PQ\)-এর মধ্যবিন্দু \(\therefore R\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
\(\Rightarrow E(\frac{2}{2}, \frac{2}{2})\)
\(\Rightarrow E(1, 1)\)
\(BD\) এর মধ্যবিন্দু \(E(\frac{3+x}{2}, \frac{4+y}{2})\) ➜Note \(\because \) সামান্তরিকের কর্ণদ্বয় পরস্পরকে সমদ্বিখণ্ডিত করে
\(\therefore E(\frac{3+x}{2}, \frac{4+y}{2})\Rightarrow E(1, 1)\)
\(\Rightarrow \frac{3+x}{2}=1, \frac{4+y}{2}=1\)
\(\Rightarrow 3+x=1\times 2, 4+y=1\times 2\)
\(\Rightarrow 3+x=2, 4+y=2\)
\(\Rightarrow x=2-3, y=2-4\)
\(\Rightarrow x=-1, y=-2\)
\(\therefore D(-1, -2)\)
\((b)\) \(AC\) ও \(BD\) এর ছেদবিন্দু \(E(1, 1)\)
\((c)\) \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}1 \ \ \ \ 3 \ \ \ \ 1 \ \ -1 \ \ \ \ 1\\2 \ \ \ \ 4 \ \ \ \ 0 \ \ -2 \ \ \ \ 2\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দু চারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{1.4-3.2+3.0-1.4+1.(-2)-(-1).0+(-1).2-1.(-2)\}\)
\(=\frac{1}{2}(4-6+0-4-2+0-2+2)\)
\(=\frac{1}{2}(6-14)\)
\(=\frac{1}{2}\times -8\)
\(=-4\)
\(\therefore \Box ABCD=4\) বর্গ একক। ➜Note \(\because \Box \neq -ve\)

দেওয়া আছে, \(A(2, 3)\), \(B(-3, 6)\), \(C(0, -5)\) এবং \(D(4, -7)\).
\(\therefore \Box ABCD=\frac{1}{2}\left|\begin{array}{c}2 \ \ -3 \ \ \ \ 0 \ \ \ \ 4 \ \ \ \ 2\\3 \ \ \ \ 6 \ \ -5 \ \ -7 \ \ \ \ 3\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দু চারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{2.6-(-3).3+(-3).(-5)-0.6+0.(-7)-4.(-5)+4.3-2.(-7)\}\)
\(=\frac{1}{2}(12+9+15-0-0+20+12+14)\)
\(=\frac{1}{2}\times 82\)
\(=41\)
\(\therefore \Box ABCD=41\) বর্গ একক।

দেওয়া আছে, \(A(1, 2)\), \(B(-5, 6)\), \(C(7, -4)\) এবং \(D(k, 2)\).
শর্তমতে,
\(\Box ABCD=12\)
\(\therefore \frac{1}{2}\left|\begin{array}{c}1 \ \ -5 \ \ \ \ \ 7 \ \ \ \ \ k \ \ \ \ \ 1\\2 \ \ \ \ \ 6 \ \ -4 \ \ \ \ 2 \ \ \ \ 2\end{array}\right|=12\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\), \(C(x_{3}, y_{3})\) এবং \(D(x_{4}, y_{4})\) বিন্দু চারটি \(\Box ABCD\) এর শীর্ষবিন্দু হলে, \(\Box ABCD=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3} \ \ x_{4} \ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{4} \ \ y_{1}\end{array}\right|\)
\(\Rightarrow \frac{1}{2}\{1.6-(-5).2+(-5).(-4)-7.6+7.2-k.(-4)+k.2-1.2\}=12\)
\(\Rightarrow (6+10+20-42+14+4k+2k-2)=12\times 2\)
\(\Rightarrow 6k+50-44=24\)
\(\Rightarrow 6k+6=24\)
\(\Rightarrow 6k=24-6\)
\(\Rightarrow k=\frac{18}{6}\)
\(\therefore k=3\)

\((a)\) দেওয়া আছে, \(A(2, 5)\), \(B(5, 9)\) এবং \(D(6, 8)\).
\(\therefore \triangle ABD=\frac{1}{2}\left|\begin{array}{c}2 \ \ \ \ 5 \ \ \ \ \ 6 \ \ \ \ \ 2\\5 \ \ \ \ 9 \ \ \ \ 8 \ \ \ \ 5\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}(2.9-5.5+5.8-6.9+6.5-2.8)\)
\(=\frac{1}{2}(18-25+40-54+30-16)\)
\(=\frac{1}{2}(88-95)\)
\(=\frac{1}{2}\times -7\)
\(=-\frac{7}{2}\)
\(=-3.5\)
\(\therefore \triangle ABD=3.5\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)
\((b)\) দেওয়া আছে, \(A(2, 5)\), \(B(5, 9)\) এবং \(D(6, 8)\).
ধরি, চতুর্থ শীর্ষবিন্দু \(C(x, y)\)
\(AC\) এর মধ্যবিন্দু \(E(\frac{2+x}{2}, \frac{5+y}{2})\)
আবার,
\(BD\) এর মধ্যবিন্দু \(E(\frac{5+6}{2}, \frac{9+8}{2})\)
\(\Rightarrow E(\frac{11}{2}, \frac{17}{2})\)
এখন,
\(E(\frac{2+x}{2}, \frac{5+y}{2})\Rightarrow E(\frac{11}{2}, \frac{17}{2})\) ➜Note রম্বসের কর্ণদ্বয় পরস্পরকে সমকোণে সমদ্বিখণ্ডিত করে।
\(\Rightarrow \frac{2+x}{2}=\frac{11}{2}, \frac{5+y}{2})=\frac{17}{2}\)
\(\Rightarrow 2+x=11, 5+y=17\)
\(\Rightarrow x=11-2, y=17-5\)
\(\Rightarrow x=9, y=12\)
\(\therefore C(9, 12)\)
[ দেখানো হলো ]
\((c)\) কর্ণ \(AC=\sqrt{(2-9)^{2}+(5-12)^{2}}\)
\(=\sqrt{(-7)^{2}+(-7)^{2}}\)
\(=\sqrt{49+49}\)
\(=\sqrt{2\times 49}\)
\(=\sqrt{2\times 7^{2}}\)
\(=7\sqrt{2}\)
কর্ণ \(BD=\sqrt{(5-6)^{2}+(9-8)^{2}}\)
\(=\sqrt{(-1)^{2}+1^{2}}\)
\(=\sqrt{1+1}\)
\(=\sqrt{2}\)
এখন,
\(\Box ABCD=\frac{1}{2}\times AC\times BD\)
\(=\frac{1}{2}\times 7\sqrt{2}\times \sqrt{2}\) ➜Note \(\because AC=7\sqrt{2}, BD=\sqrt{2}\)
\(=\frac{1}{2}\times 7\times 2\)
\(=\frac{1}{2}\times 14\)
\(=7\) বর্গ একক।

\((a)\) দেওয়া আছে, \(A(0, -1)\), \(B(15, 2)\), \(C(-1, 2)\) এবং \(D(4, -5)\).
\(\therefore AB=\sqrt{(0-15)^{2}+(-1-2)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-15)^{2}+(-3)^{2}}\)
\(=\sqrt{225+9}\)
\(=\sqrt{234}\)
\(=\sqrt{26\times 9}\)
\(=\sqrt{26\times 3^{2}}\)
\(=3\sqrt{26}\)
\(=3\sqrt{13\times 2}\)
\(=3\sqrt{13}\times\sqrt{2}\)
আবার,
\(CD=\sqrt{(-1-4)^{2}+(2+5)^{2}}\) ➜Note\(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-5)^{2}+(7)^{2}}\)
\(=\sqrt{25+49}\)
\(=\sqrt{74}\)
\(=\sqrt{37\times 2}\)
\(=\sqrt{37}\times\sqrt{2}\)
এখন,
\(\frac{AB}{CD}=\frac{3\sqrt{13}\times\sqrt{2}}{\sqrt{37}\times\sqrt{2}}\)
\(\Rightarrow \frac{AB}{CD}=\frac{3\sqrt{13}}{\sqrt{37}}\)
\(\therefore AB:CD=3\sqrt{13}:\sqrt{37}\)
\((b)\) \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}\ 0 \ \ \ \ 15 \ -1 \ \ \ \ \ 0\\-1 \ \ \ \ 2 \ \ \ \ \ 2 \ \ \ -1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}[0.2-15.(-1)+15.2-(-1).2+(-1).(-1)-0.2]\)
\(=\frac{1}{2}(0+15+30+2+1-0)\)
\(=\frac{1}{2}\times 48\)
\(=24\)
\(\therefore \triangle ABC=24\) বর্গ একক।
\(\triangle ABD=\frac{1}{2}\left|\begin{array}{c}\ \ 0 \ \ \ 15 \ \ \ \ \ \ 4 \ \ \ \ \ \ \ 0\\-1 \ \ \ \ 2 \ \ -5 \ \ -1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=\frac{1}{2}\{0.2-15.(-1)+15.(-5)-4.2+4.(-1)-0.(-5)\} \)
\(=\frac{1}{2}(0+15-75-8-4+0)\)
\(=\frac{1}{2}(15-87)\)
\(=\frac{1}{2}\times -72\)
\(=-36\)
\(\therefore \triangle ABD=36\) বর্গ একক। ➜Note \(\because \triangle \neq -ve\)
এখন,
\(\frac{\triangle ABC}{\triangle ABD}=\frac{24}{36}\)
\(\Rightarrow \frac{\triangle ABC}{\triangle ABD}=\frac{2}{3}\)
\(\therefore \triangle ABC:\triangle ABD=2:3\).
\((a)\) এখন,
\(\delta_{ABC}=\left|\begin{array}{c}\ 0 \ \ \ \ 15 \ -1 \ \ \ \ \ 0\\-1 \ \ \ \ 2 \ \ \ \ \ 2 \ \ \ -1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=0.2-15.(-1)+15.2-(-1).2+(-1).(-1)-0.2\)
\(=0+15+30+2+1-0\)
\(=48\)
আবার,
\(\delta_{ABD}=\left|\begin{array}{c}\ \ 0 \ \ \ 15 \ \ \ \ \ \ 4 \ \ \ \ \ \ \ 0\\-1 \ \ \ \ 2 \ \ -5 \ \ -1\end{array}\right|\) ➜Note কোন সমতলে \(A(x_{1}, y_{1})\), \(B(x_{2}, y_{2})\) এবং \(C(x_{3}, y_{3})\) বিন্দুতিনটি \(\triangle ABC\) এর শীর্ষবিন্দু হলে, \(\triangle ABC=\frac{1}{2}\left|\begin{array}{c}x_{1} \ \ x_{2} \ \ x_{3}\ \ x_{1}\\y_{1} \ \ y_{2} \ \ y_{3} \ \ y_{1}\end{array}\right|\)
\(=0.2-15.(-1)+15.(-5)-4.2+4.(-1)-0.(-5)\)
\(=0+15-75-8-4+0\)
\(=15-87\)
\(=-72\)
আবার,
\(\therefore \frac{\delta_{ABC}}{\delta_{ABD}}=\frac{48}{-72}\)
\(\Rightarrow \frac{\delta_{ABC}}{\delta_{ABD}}=-\frac{2}{3}\)
\(\Rightarrow \delta_{ABC}:\delta_{ABD}=-2:3\)
\(\therefore CD\) সরলরেখা \(AB\) সরলরেখাকে \(2:3\) অনুপাতে অন্তর্বিভক্ত করে। [\(\because\) অনুপাত ঋণাত্মক \((-ve)\)]