মনে করি,
\(P(x_{1}, y_{1})\) ও \(Q(x_{2}, y_{2})\) যে কোন দুইটি বিন্দু।
\(P\) এবং \(Q\) হতে \(OX\) এর উপর যথাক্রমে \(PM\) এবং \(QN\) লম্ব অংকন করি।
তাহলে, \(OM=x_{1}\), \(PM=y_{1}\), \(ON=x_{2}\), \(QN=y_{2}\). আবার,
\(QR=NM=OM-ON=x_{1}-x_{2}\).
\(PR=PM-RM=y_{1}-y_{2}\).
\(PQR\) সমকোণী ত্রীভুজ হতে পাই,
\(PQ^{2}=QR^{2}+PR^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\)
\(\Rightarrow PQ^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\)
\(\therefore PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\) ➜Note \(\because\) দূরত্ব সর্বদা ধনাত্মক।
\(PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)

মনে করি,
\(P(x_{1}, y_{1})\) ও \(Q(x_{2}, y_{2})\) যে কোন দুইটি বিন্দু। মূলবিন্দু বা \(O\) বিন্দুর সাপেক্ষে বিন্দুদ্বয়ের অবস্থান ভেক্টর যথাক্রমে \(\overline{OP}=x_{1}\hat{i}+y_{1}\hat{j}\) এবং \(\overline{OQ}=x_{2}\hat{i}+y_{2}\hat{j}\)।
\(\triangle OPQ\) হতে পাই,
\(\overline{PQ}=\overline{OQ}-\overline{OP}\)
\(=x_{2}\hat{i}+y_{2}\hat{j}-x_{1}\hat{i}-y_{1}\hat{j}\)
\(=(x_{2}-x_{1})\hat{i}+(y_{2}-y_{1})\hat{j}\)
\(P(x_{1}, y_{1})\) ও \(P(x_{2}, y_{2})\) বিন্দুর দূরত্ব \(\overline{\mid PQ \mid}\)
\(=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)
\(=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)
\(PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)
উদাহরণঃ \(P(1, 2)\) ও \(Q(-2, 6)\) বিন্দু দ্বয়ের দূরত্ব
\(=\sqrt{(1+2)^{2}+(2-6)^{2}}\)
\(=\sqrt{3^{2}+4^{2}}\) \(=\sqrt{9+19}\)
\(=\sqrt{25}\)
\(=5\)।

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 \(ABC\) ত্রিভুজের শীর্ষত্রয় \(A(0, -4)\), \(B(0, 4)\) এবং \(C(x, y)\)
শর্তমতে,
\(AB=BC=AC\)
\(\Rightarrow AB^{2}=BC^{2}=AC^{2}\)
এখন,
\(\Rightarrow AB^{2}=BC^{2}\)
\(\Rightarrow (0-0)^{2}+(-4-4)^{2}=(0-x)^{2}+(4-y)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow 0+(-8)^{2}=x^{2}+4^{2}-2.4.y+y^{2}\)
\(\Rightarrow 64=x^{2}+16-8y+y^{2}\)
\(\Rightarrow x^{2}-8y+y^{2}-48=0.......(i)\)
আবার,
\(\Rightarrow AB^{2}=AC^{2}\)
\(\Rightarrow (0-0)^{2}+(-4-4)^{2}=(0-x)^{2}+(-4-y)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow (-8)^{2}=x^{2}+(4+y)^{2}\)
\(\Rightarrow 64=x^{2}+4^{2}+2.4.y+y^{2}\)
\(\Rightarrow 64=x^{2}+16+8y+y^{2}\)
\(\Rightarrow x^{2}+8y+y^{2}-48=0.......(ii)\)
\((ii)-(i)\) এর সাহায্যে, \( x^{2}+8y+y^{2}-48-x^{2}+8y-y^{2}+48=0\)
\(\Rightarrow 16y=0\) \(\therefore y=0\)
\(y\)-এর মান \((i)\) সমীকরনে বসিয়ে পাই,
\(x^{2}-48=0\)
\(\Rightarrow x^{2}=48\)
\(\Rightarrow x=\pm\sqrt{48}\)
\(\Rightarrow x=\pm\sqrt{3\times 4^{2}}\)
\(\Rightarrow x=\pm 4\sqrt{3}\)
\(\therefore \) তৃতীয় শীর্ষবিন্দুর স্থানাঙ্ক \((4\sqrt{3}, 0)\) বা, \((-4\sqrt{3}, 0)\)

মনে করি, \(A(1, 2)\), \(B(-3, 1)\), \(C(-2, -3)\) এবং \(D(2, -2)\)
এখন,
\(AB=\sqrt{(1+3)^{2}+(2-1)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{4^{2}+1^{2}}\)
\(=\sqrt{16+1}\)
\(=\sqrt{17}\)
\(BC=\sqrt{(-3+2)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{1^{2}+4^{2}}\)
\(=\sqrt{1+16}\)
\(=\sqrt{17}\)
\(CD=\sqrt{(-2-2)^{2}+(-3+2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-4)^{2}+(-1)^{2}}\)
\(=\sqrt{16+1}\)
\(=\sqrt{17}\)
\(DA=\sqrt{(2-1)^{2}+(-2-2)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{1^{2}+(-4)^{2}}\)
\(=\sqrt{1+16}\)
\(=\sqrt{17}\)
এখানে, \(AB=BC=CD=DA\)
\(\therefore ABCD\) চতুর্ভুজটি বর্গক্ষেত্র বা, রম্বস হতে পারে।
আবার,
\(AC=\sqrt{(1+2)^{2}+(2+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{3^{2}+5^{2}}\)
\(=\sqrt{9+25}\)
\(=\sqrt{34}\)
\(BD=\sqrt{(-3-2)^{2}+(1+2)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-5)^{2}+3^{2}}\)
\(=\sqrt{25+9}\)
\(=\sqrt{34}\)
\(\because AB=BC=CD=DA\) এবং \(AC=BD\)
\(\therefore ABCD\) চতুর্ভুজটি একটি বর্গক্ষেত্র।

দেওয়া আছে, \(A(3, -5)\), \(B(9, 10)\), \(C(3, 25)\) এবং \(D(-3, 10)\)
এখন,
\(AB=\sqrt{(3-9)^{2}+(-5-10)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-6)^{2}+(-15)^{2}}\)
\(=\sqrt{36+225}\)
\(=\sqrt{261}\)
\(BC=\sqrt{(9-3)^{2}+(10-25)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{6^{2}+(-15)^{2}}\)
\(=\sqrt{36+225}\)
\(=\sqrt{261}\)
\(CD=\sqrt{(3+3)^{2}+(25-10)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{6^{2}+15^{2}}\)
\(=\sqrt{36+225}\)
\(=\sqrt{261}\)
\(DA=\sqrt{(-3-3)^{2}+(10+5)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-6)^{2}+15^{2}}\)
\(=\sqrt{36+225}\)
\(=\sqrt{261}\)
এখানে, \(AB=BC=CD=DA\)
\(\therefore ABCD\) চতুর্ভুজটি বর্গক্ষেত্র বা, রম্বস হতে পারে।
আবার,
\(AC=\sqrt{(3-3)^{2}+(-5-25)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{0^{2}+(-30)^{2}}\)
\(=\sqrt{0+900}\)
\(=\sqrt{900}\)
\(=30\)
\(BD=\sqrt{(9+3)^{2}+(10-10)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{12^{2}+0^{2}}\)
\(=\sqrt{144+0}\)
\(=\sqrt{144}\)
\(=12\)
\(\because AB=BC=CD=DA\) এবং \(AC \neq BD\)
\(\therefore ABCD\) চতুর্ভুজটি একটি রম্বস।

মনে করি, \(P(4, k)\) এবং \(N(-5, -7)\).
এখন,
X-অক্ষ থেকে \((4, K)\) বিন্দুর দূরত্ব \(PM=\mid k \mid\)
\(P(4, k)\) থেকে \(N(-5, -7)\) এর দূরত্ব,
\(PN=\sqrt{(4+5)^{2}+(k+7)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{9^{2}+(k)^{2}+2.k.7+7^{2}}\)
\(=\sqrt{81+(k)^{2}+14k+49}\)
\(=\sqrt{(k)^{2}+14k+130}\)
শর্তমতে, \(PN=PM\)
\(\Rightarrow \sqrt{(k)^{2}+14k+130}=\mid k \mid\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow (k)^{2}+14k+130=k^{2}\)
\(\Rightarrow (k)^{2}+14k+130-k^{2}=0\)
\(\Rightarrow 14k+130=0\)
\(\Rightarrow 14k=-130\)
\(\Rightarrow k=\frac{-130}{14}\)
\(\therefore k=-\frac{65}{7}\).

মনে করি, \(O(0, 0)\), \(A(0, 8)\), \(B(4, 0)\) এবং পরিকেন্দ্র \(C(x, y)\).
শর্তমতে,
\(CO=CA=CB\)
\(\Rightarrow CO=CA\)
\(\Rightarrow CO^{2}=CA^{2}\)
\(\Rightarrow (x-0)^{2}+(y-0)^{2}=(x-0)^{2}+(y-8)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2.y.8+8^{2}\)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-16y+64\)
\(\Rightarrow x^{2}+y^{2}-x^{2}-y^{2}+16y-64=0\)
\(\Rightarrow 16y-64=0\)
\(\Rightarrow 16y=64\)
\(\Rightarrow y=\frac{64}{16}\)
\(\therefore y=4\)
আবার,
\(\Rightarrow CO=CB\)
\(\Rightarrow CO^{2}=CB^{2}\)
\(\Rightarrow (x-0)^{2}+(y-0)^{2}=(x-4)^{2}+(y-0)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}+y^{2}=x^{2}-2.x.4+4^{2}+(y)^{2}\)
\(\Rightarrow x^{2}+y^{2}=x^{2}-8x+16+y^{2}\)
\(\Rightarrow x^{2}+y^{2}-x^{2}-y^{2}+8x-16=0\)
\(\Rightarrow 8x-16=0\)
\(\Rightarrow 8x=16\)
\(\Rightarrow x=\frac{16}{8}\)
\(\therefore x=2\)
\(\therefore \triangle OAB\) এর পরিকেন্দ্র \(C(2, 4)\)

মনে করি, বৃত্তের কেন্দ্র \(C(5, 3)\) এবং \(AB\) জ্যা-এর মধ্যবিন্দু \(D(3, 2)\)
শর্তমতে, বৃত্তের ব্যাসার্ধ \(CA=CB=5\)
\(C,D\) যোগ করি। তাহলে, \(CD \bot AB\) ➜Note বৃত্তের কেন্দ্র হতে জ্যা-এর মধ্যবিন্দুর সংযোগ সরলরেখা জ্যা-এর উপর লম্ব।
এখন, \(CD^{2}=(5-3)^{2}+(3-2)^{2}\)
\(=2^{2}+1^{2}\)
\(=4+1\)
\(=5\)
\(CAD\) সমকোণী ত্রিভুজ হতে, \(CA^{2}=AD^{2}+CD^{2}\)
\(\Rightarrow 5^{2}=AD^{2}+5\) ➜Note\(\because CA=5, CD^{2}=5\)
\(\Rightarrow 25=AD^{2}+5\)
\(\Rightarrow AD^{2}+5=25\)
\(\Rightarrow AD^{2}=25-5\)
\(\Rightarrow AD^{2}=20\)
\(\Rightarrow AD=\sqrt{20}\)
\(\Rightarrow AD=\sqrt{5\times 2^{2}}\)
\(\therefore AD=2\sqrt{5}\)
আবার,
জ্যা-এর দৈর্ঘ্য \(AB=2\times AD\)
\(=2\times 2\sqrt{5}\)
\(=4\sqrt{5}\) ➜Note\(\because AD=2\sqrt{5}\)
\(\therefore \) জ্যা-এর দৈর্ঘ্য \(=4\sqrt{5}\).

মনে করি, \(A(1, 2)\), \(B(3, 4)\) এবং \(C(6, 7)\)
এখানে,
\(AB=\sqrt{(1-3)^{2}+(2-4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-2)^{2}+(-2)^{2}}\)
\(\Rightarrow AB=\sqrt{4+4}\)
\(\Rightarrow AB=\sqrt{8}\)
\(\Rightarrow AB=\sqrt{2\times 2^{2}}\)
\(\therefore AB=2\sqrt{2}\)
আবার,
\(BC=\sqrt{(3-6)^{2}+(4-7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-3)^{2}+(-3)^{2}}\)
\(\Rightarrow BC=\sqrt{9+9}\)
\(\Rightarrow BC=\sqrt{18}\)
\(\Rightarrow BC=\sqrt{2\times 3^{2}}\)
\(\therefore BC=3\sqrt{2}\)
আবার,
\(AC=\sqrt{(1-6)^{2}+(2-7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-5)^{2}+(-5)^{2}}\)
\(\Rightarrow AC=\sqrt{25+25}\)
\(\Rightarrow AC=\sqrt{50}\)
\(\Rightarrow AC=\sqrt{2\times 5^{2}}\)
\(\therefore AC=5\sqrt{2}\)
এখানে, \(AB+BC=2\sqrt{2}+3\sqrt{2}\)
\(=5\sqrt{2}\)
\(=AC\) ➜Note\(\because AB=2\sqrt{2}, BC=3\sqrt{2}, AC=5\sqrt{2} \)
\(\because AB+BC=AC\)
\(\therefore A, B, C\) বিন্দুত্রয় একই সরলরেখার উপর অবস্থিত।

মনে করি, বিন্দুটির ভুজ \(x\)
\(\therefore \) বিন্দুটির স্থানাঙ্ক \(P(x, 6)\) এবং অপর বিন্দুটি \(Q(5, 6)\)
শর্তমতে, \(\mid PQ \mid=4\)
\(\Rightarrow PQ^{2}=4^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-5)^{2}+(6-6)^{2}=16\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow (x-5)^{2}+0^{2}=16\)
\(\Rightarrow (x-5)=\pm\sqrt{16}\)
\(\Rightarrow (x-5)=\pm 4\)
\(\Rightarrow x=\pm 4+5\)
\(\Rightarrow x=4+5, -4+5\)
\(\Rightarrow x=9, 1\)
\(\therefore \) বিন্দুটির ভুজ \(=9, 1\).

মনে করি, বিন্দুটি \(A(a, 5)\),
\(Y\) অক্ষ হতে \(A(a, 5)\) এর দূরত্ব \(=a\)।
শর্তমতে, \(\mid AP \mid=a\)
\(\Rightarrow AP^{2}=a^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (a-7)^{2}+(5-2)^{2}=a^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow a^{2}-2.a.7+7^{2}+3^{2}=a^{2}\)
\(\Rightarrow a^{2}-14a+49+9=a^{2}\)
\(\Rightarrow a^{2}-14a+58-a^{2}=0\)
\(\Rightarrow -14a+58=0\)
\(\Rightarrow -14a=-58\)
\(\Rightarrow a=\frac{-58}{-14}\)
\(\Rightarrow a=\frac{29}{7}\)
\(\therefore a=\frac{29}{7}\).

মনে করি, বিন্দুটি \(A(4, k)\),
\(X\) অক্ষ হতে \(A(4, k)\) এর দূরত্ব \(=k\)।
দেওয়া আছে,\(P(-5, -7)\)
শর্তমতে, \(\mid AP \mid=k\)
\(\Rightarrow AP^{2}=k^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (4+5)^{2}+(k+7)^{2}=k^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow 9^{2}+k^{2}+2.k.7+7^{2}=k^{2}\)
\(\Rightarrow 81+k^{2}+14k+49=k^{2}\)
\(\Rightarrow k^{2}+14k+130-k^{2}=0\)
\(\Rightarrow 14k+130=0\)
\(\Rightarrow 14k=-130\)
\(\Rightarrow k=\frac{-130}{14}\)
\(\Rightarrow k=-\frac{65}{7}\)
\(\therefore k=-\frac{65}{7}\).

মনে করি,\(A(4, 2)\),\(B(7, 5)\) এবং \(C(9, 7)\).
এখন,
\(AB=\sqrt{(4-7)^{2}+(2-5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-3)^{2}+(-3)^{2}}\)
\(=\sqrt{9+9}\)
\(=\sqrt{18}\)
\(=\sqrt{2\times 3^{2}}\)
\(=3\sqrt{2}\)
\(\therefore AB=3\sqrt{2}\)
আবার,
\(BC=\sqrt{(7-9)^{2}+(5-7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-2)^{2}+(-2)^{2}}\)
\(=\sqrt{4+4}\)
\(=\sqrt{8}\)
\(=\sqrt{2\times 2^{2}}\)
\(=2\sqrt{2}\)
\(\therefore BC=2\sqrt{2}\)
আবার,
\(AC=\sqrt{(4-9)^{2}+(2-7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-5)^{2}+(-5)^{2}}\)
\(=\sqrt{25+25}\)
\(=\sqrt{50}\)
\(=\sqrt{2\times 5^{2}}\)
\(=5\sqrt{2}\)
\(\therefore AC=5\sqrt{2}\)
এখন, \(AB+BC=3\sqrt{2}+2\sqrt{2}\) \(=5\sqrt{2}\) \(=AC\)
\(\Rightarrow AB+BC=AC\)
\(\because AB+BC=AC\)
\(\therefore \) বিন্দু তিনটি একই সরলরেখায় অবস্থিত।

মনে করি,\(A(-5, 3)\), \(B(15, -9)\) এবং মূলবিন্দু \(O(0, 0)\).
এখন,
\(AO=\sqrt{(-5-0)^{2}+(3-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-5)^{2}+3^{2}}\)
\(=\sqrt{25+9}\)
\(=\sqrt{34}\)
\(\therefore AO=\sqrt{34}\)
আবার,
\(BO=\sqrt{(15-0)^{2}+(-9-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-15)^{2}+(-9)^{2}}\)
\(=\sqrt{225+81}\)
\(=\sqrt{306}\)
\(=\sqrt{34\times 9}\)
\(=\sqrt{34\times 3^{2}}\)
\(=3\sqrt{34}\)
\(\therefore BO=3\sqrt{34}\)
আবার,
\(AB=\sqrt{(-5-15)^{2}+(3+9)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-20)^{2}+12^{2}}\)
\(=\sqrt{400+144}\)
\(=\sqrt{544}\)
\(=\sqrt{34\times 16}\)
\(=\sqrt{34\times 4^{2}}\)
\(=4\sqrt{34}\)
\(\therefore AB=4\sqrt{34}\)
এখন,
\(AO+BO=\sqrt{34}+3\sqrt{34}=4\sqrt{34}=AB\)
\(\Rightarrow AO+BO=AB\)
\(\because AO+BO=AB\)
\(\therefore \) বিন্দু দুইটির সংযোগরেখা মূলবিন্দু দিয়ে যায়।

মনে করি, \(P(x, y)\).
দেওয়া আছে, \(A(1, -1)\),\(B(9, 7)\), \(C(1, 7)\).
শর্তমতে, \(PA=PB=PC\)
\(\Rightarrow PA=PB\)
\(\Rightarrow PA^{2}=PB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-1)^{2}+(y+1)^{2}=(x-9)^{2}+(y-7)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.1+1^{2}+y^{2}+2.y.1+1^{2}=x^{2}-2.x.9+9^{2}+y^{2}-2.y.7+7^{2}\)
\(\Rightarrow x^{2}-2x+1+y^{2}+2y+1=x^{2}-18x+81+y^{2}-14y+49\)
\(\Rightarrow x^{2}-2x+1+y^{2}+2y+1-x^{2}+18x-81-y^{2}+14y-49=0\)
\(\Rightarrow 16x+16y-128=0\)
\(\Rightarrow x+y-8=0\)
\(\Rightarrow x=8-y ......(i)\)
আবার,
\(\Rightarrow PA=PC\)
\(\Rightarrow PA^{2}=PB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-1)^{2}+(y+1)^{2}=(x-1)^{2}+(y-7)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.1+1^{2}+y^{2}+2.y.1+1^{2}=x^{2}-2.x.1+1^{2}+y^{2}-2.y.7+7^{2}\)
\(\Rightarrow x^{2}-2x+1+y^{2}+2y+1=x^{2}-2x+1+y^{2}-14y+49\)
\(\Rightarrow x^{2}-2x+1+y^{2}+2y+1-x^{2}+2x-1-y^{2}+14y-49=0\)
\(\Rightarrow 16y-48=0\)
\(\Rightarrow 16y=48\)
\(\Rightarrow y=\frac{48}{16}\)
\(\therefore y=3\)
\(y=3\), \((i)\) -এ বসিয়ে পাই।
\(\Rightarrow x=8-3\)
\(\therefore x=5\)
\(\therefore \) \(P\) বিন্দুর স্থানাঙ্ক \((5, 3)\).

শর্তমতে, \(P(x, 0)\) ➜Note \(X\) অক্ষের উপর অবস্থিত যে কোন বিন্দুর \(y\) স্থানাঙ্ক \(0\) হয়।
দেওয়া আছে, \(A(1, -1)\),\(B(9, 7)\).
আবার,
শর্তমতে, \(PA=PB\)
\(\Rightarrow PA^{2}=PB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-1)^{2}+(0+1)^{2}=(x-9)^{2}+(0-7)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.1+1^{2}+1^{2}=x^{2}-2.x.9+9^{2}+(-7)^{2}\)
\(\Rightarrow x^{2}-2x+1+1=x^{2}-18x+81+49\)
\(\Rightarrow x^{2}-2x+1+1-x^{2}+18x-81-49=0\)
\(\Rightarrow 16x-128=0\)
\(\Rightarrow 16x=128\)
\(\Rightarrow x=\frac{128}{16}\)
\(\Rightarrow x=8\)
\(\therefore \) \(P\) বিন্দুর স্থানাঙ্ক \((8, 0)\).

মনে করি,\(P(x, y)\), \(A(3, 5)\) এবং \(B(-6, -9)\).
শর্তমতে, \(PA=PB\)
\(\Rightarrow PA^{2}=PB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-3)^{2}+(y-5)^{2}=(x+6)^{2}+(y+9)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.3+3^{2}+y^{2}-2.y.5+5^{2}=x^{2}+2.x.6+6^{2}+y^{2}+2.y.9+9^{2}\)
\(\Rightarrow x^{2}-6x+9+y^{2}-10y+25=x^{2}+12x+36+y^{2}+18y+81\)
\(\Rightarrow x^{2}-6x+9+y^{2}-10y+25-x^{2}-12x-36-y^{2}-18y-81=0\)
\(\Rightarrow -6x+9-10y+25-12x-36-18y-81=0\)
\(\Rightarrow -18x-28y-83=0\)
\(\Rightarrow 18x+28y+83=0\) ➜Note উভয় পার্শে \(-1\) গুণ করে।
\(\therefore 18x+28y+83=0\)
[ প্রমাণিত ]

মনে করি,
\(P(x, y)\), \(A(a+b, b-a)\) এবং \(B(a-b, a+b)\).
শর্তমতে, \(PA=PB\)
\(\Rightarrow PA^{2}=PB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}+a^{2}+b^{2}-2.x.a+2.a.b-2.x.b+y^{2}+a^{2}+b^{2}-2.y.b-2.b.a+2.y.a\)
\(=x^{2}+a^{2}+b^{2}-2.x.a-2.a.b+2.x.b+y^{2}+a^{2}+b^{2}-2.y.a+2.a.b-2.y.b\)
\(\Rightarrow x^{2}+a^{2}+b^{2}-2.x.a+2.a.b-2.x.b+y^{2}+a^{2}+b^{2}-2.y.b-2.b.a+2.y.a\)
\(-x^{2}-a^{2}-b^{2}+2.x.a+2.a.b-2.x.b-y^{2}-a^{2}-b^{2}+2.y.a-2.a.b+2.y.b=0\)
\(\Rightarrow -2ax+2ab-2bx-2by-2ab+2ay+2ax+2ab-2bx+2ay-2ab+2by=0\)
\(\Rightarrow -2bx+2ay-2bx+2ay=0\)
\(\Rightarrow -4bx+4ay=0\)
\(\Rightarrow -4bx=-4ay\)
\(\Rightarrow bx=ay\) ➜Note উভয় পার্শে \(-4\) ভাগ করে।
\(\therefore bx=ay\)
[ প্রমাণিত ]

মনে করি,
বিন্দুটির ভুজ=\(x\)
শর্তমতে, কটি= \(2x\)।
\(\therefore \) বিন্দুটি \(P(x, 2x)\)।
এবং \(Q(4, 3)\)
এখন, \(PQ=\sqrt{10}\)
\(\Rightarrow PQ^{2}=10\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-4)^{2}+(2x-3)^{2}=10\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.4+4^{2}+(2x)^{2}-2.2x.3+3^{2}=10\)
\(\Rightarrow x^{2}-8x+16+4x^{2}-12x+9-10=0\)
\(\Rightarrow 5x^{2}-20x+15=0\)
\(\Rightarrow x^{2}-4x+3=0\) ➜Note উভয় পার্শে \(5\) ভাগ করে।
\(\Rightarrow x^{2}-3x-x+3=0\)
\(\Rightarrow x(x-3)-1(x-3)=0\)
\(\Rightarrow (x-3)(x-1)=0\)
\(\Rightarrow (x-3)=0,(x-1)=0\)
\(\Rightarrow x=3, x=1\)
\(\therefore\) বিন্দুটির স্থানাঙ্ক \(P(3, 6)\) অথবা, \(P(1, 2)\).

দেওয়া আছা,
\(P(-7, -1)\), \(Q(-3, 2)\), \(R(x, 5)\) এবং \(PQ=QR\)
\(\Rightarrow PQ^{2}=QR^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (-7+3)^{2}+(-1-2)^{2}=(-3-x)^{2}+(2-5)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow (-4)^{2}+(-3)^{2}=(x+3)^{2}+(-3)^{2}\)
\(\Rightarrow 16+9=(x+3)^{2}+9\)
\(\Rightarrow (x+3)^{2}+9=16+9\)
\(\Rightarrow (x+3)^{2}=16+9-9\)
\(\Rightarrow (x+3)^{2}=16\)
\(\Rightarrow x+3=\pm\sqrt{16}\)
\(\Rightarrow x=\pm 4-3\)
\(\Rightarrow x= 4-3, -4-3\)
\(\Rightarrow x= 1, -7\)
\(\therefore \) \(x\) এর মান \( 1\) অথবা, \(-7\)

মনে করি, \(A(2\sqrt{3}, 90^{o})\),\(B(2, 120^{o})\) এবং \(C(2, 60^{o})\)
এখানে, \((r_{1}, \theta_{1}) \) \(\Rightarrow \) \(A(2\sqrt{3}, 90^{o})\)
\(\therefore \) \(r_{1}=2\sqrt{3}, \theta_{1}=90^{o} \)
আবার,
\((r_{2}, \theta_{2}) \) \(\Rightarrow \) \(B(2, 120^{o})\)
\(\therefore \) \(r_{2}=2, \theta_{2}=120^{o} \)
তাহলে,
\(AB=\sqrt{(2\sqrt{3})^{2}+2^{2}-2.2\sqrt{3}.2\cos(90^{o}-120^{o})}\) ➜Note\(\because PQ=\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}\)
\(\Rightarrow AB=\sqrt{12+4-8\sqrt{3}\cos(-30^{o})}\)
\(\Rightarrow AB=\sqrt{16-8\sqrt{3}\cos30^{o}}\) ➜Note \(\because \cos(-\theta)=\cos\theta\)
\(\Rightarrow AB=\sqrt{16-8\sqrt{3}\times \frac{\sqrt{3}}{2}}\) ➜Note\(\because \cos30^{o}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow AB=\sqrt{16-4\times 3}\)
\(\Rightarrow AB=\sqrt{16-12}\)
\(\Rightarrow AB=\sqrt{4}\)
\(\therefore AB=2\)
আবার, \((r_{1}, \theta_{1}) \) \(\Rightarrow \) \(B(2, 120^{o})\)
\(\therefore \) \(r_{1}=2, \theta_{1}=120^{o}\)
এবং
\((r_{2}, \theta_{2}) \) \(\Rightarrow \) \(C(2, 60^{o})\)
\(\therefore \) \(r_{2}=2, \theta_{2}=60^{o} \)
তাহলে,
\(\Rightarrow BC=\sqrt{2^{2}+2^{2}-2.2.2\cos(120^{o}-60^{o})}\) ➜Note\(\because PQ=\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}\)
\(\Rightarrow BC=\sqrt{4+4-8\cos60^{o}}\)
\(\Rightarrow BC=\sqrt{8-8\times \frac{1}{2}}\) ➜Note\(\because \cos60^{o}=\frac{1}{2}\)
\(\Rightarrow BC=\sqrt{8-4}\)
\(\Rightarrow BC=\sqrt{4}\)
\(\Rightarrow BC=2\)
\(\therefore BC=2\)
আবার, \((r_{1}, \theta_{1}) \) \(\Rightarrow \) \(C(2, 60^{o})\)
\(\therefore \) \(r_{1}=2, \theta_{1}=60^{o}\)
এবং
\((r_{2}, \theta_{2}) \) \(\Rightarrow \) \(A(2\sqrt{3}, 90^{o})\)
\(\therefore \) \(r_{2}=2\sqrt{3}, \theta_{2}=90^{o}\)
তাহলে,
\(\Rightarrow CA=\sqrt{2^{2}+(2\sqrt{3})^{2}-2.2.2\sqrt{3}\cos(60^{o}-90^{o})}\) ➜Note\(\because PQ=\sqrt{r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta_{1}-\theta_{2})}\)
\(\Rightarrow CA=\sqrt{4+12-8\sqrt{3}\cos(-30^{o})}\)
\(\Rightarrow CA=\sqrt{4+12-8\sqrt{3}\cos30^{o}}\) ➜Note \(\because \cos(-\theta)=\cos\theta\)
\(\Rightarrow CA=\sqrt{16-8\sqrt{3}\times \frac{\sqrt{3}}{2}}\) ➜Note \(\because \cos30^{o}=\frac{\sqrt{3}}{2}\)
\(\Rightarrow CA=\sqrt{16-4\times 3}\)
\(\Rightarrow CA=\sqrt{16-12}\)
\(\Rightarrow CA=\sqrt{4}\)
\(\Rightarrow CA=2\)
\(\therefore CA=2\)
\(\because AB=BC=CA\)
\(\therefore A, B, C\) বিন্দুগুলি একটি সমবাহু ত্রিভুজের শীর্ষবিন্দু।

মনে করি, তৃতীয় শীর্ষবিন্দু \(R(x, y)\)।
দেওয়া আছে, \(P(4, 0)\) এবং \(Q(0, 4)\)।
শর্তমতে, \(RP=RQ=PQ\) ➜Note সমবাহু ত্রিভুজের বাহুগুলি সমান।
\(\Rightarrow RP=PQ\)
\(\Rightarrow RP^{2}=PQ^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-4)^{2}+(y-0)^{2}=(4-0)^{2}+(0-4)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.4+4^{2}+y^{2}=(4)^{2}+(-4)^{2}\)
\(\Rightarrow x^{2}-8x+16+y^{2}=16+16\)
\(\Rightarrow x^{2}+y^{2}-8x+16-32=0\)
\(\Rightarrow x^{2}+y^{2}-8x-16=0 .......(i)\)
আবার,
\(\Rightarrow RQ=PQ\)
\(\Rightarrow RQ^{2}=PQ^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-0)^{2}+(y-4)^{2}=(4-0)^{2}+(0-4)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}+y^{2}-2.y.4+4^{2}=(4)^{2}+(-4)^{2}\)
\(\Rightarrow x^{2}+y^{2}-8y+16=16+16\)
\(\Rightarrow x^{2}+y^{2}-8y+16-32=0\)
\(\Rightarrow x^{2}+y^{2}-8y-16=0 .......(ii)\)
এখন, \((i)-(ii)\) এর সাহায্যে পাই,
\(x^{2}+y^{2}-8x-16-x^{2}-y^{2}+8y+16=0\)
\(\Rightarrow -8x+8y=0\)
\(\Rightarrow -8x=-8y\)
\(\Rightarrow x=y .......(iii)\) ➜Note উভয় পার্শে \(-8\) ভাগ করে।
\((iii)\) এর সাহায্য নিয়ে \((i)\) হতে পাই,
\(x^{2}+x^{2}-8x-16=0\)
\(\Rightarrow 2x^{2}-8x-16=0\)
\(\Rightarrow x^{2}-4x-8=0\)
\(\Rightarrow x=\frac{-(-4)\pm\sqrt{(-4)^{2}-4.1(-8)}}{2.1}\)
\(\Rightarrow x=\frac{4\pm\sqrt{16+32}}{2}\)
\(\Rightarrow x=\frac{4\pm\sqrt{48}}{2}\)
\(\Rightarrow x=\frac{4\pm\sqrt{3\times 4^{2}}}{2}\)
\(\Rightarrow x=\frac{4\pm 4\sqrt{3}}{2}\)
\(\Rightarrow x=\frac{2(2\pm 2\sqrt{3})}{2}\)
\(\therefore x=2\pm 2\sqrt{3}\)
এখন,
\((iii)\) হতে পাই, \( y=2\pm 2\sqrt{3}\)
\(\therefore \) তৃতীয় শীর্ষের স্থানাঙ্ক \((2+2\sqrt{3}, 2+2\sqrt{3})\), \((2-2\sqrt{3}, 2-2\sqrt{3})\)

মনে করি, \(C(x, y)\)।
দেওয়া আছে, \(A(3, 4)\) ও \(B(3, 6)\).
শর্তমতে, \(AB=BC=CA\) ➜Note সমবাহু ত্রিভুজের বাহুগুলি সমান।
\(\Rightarrow CA=AB\)
\(\Rightarrow CA^{2}=AB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-3)^{2}+(y-4)^{2}=(3-3)^{2}+(4-6)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.3+3^{2}+y^{2}-2.y.4+4^{2}=0^{2}+(-2)^{2}\)
\(\Rightarrow x^{2}-6x+9+y^{2}-8y+16=0+4\)
\(\Rightarrow x^{2}+y^{2}-6x-8y+25-4=0\)
\(\Rightarrow x^{2}+y^{2}-6x-8y+21=0 .......(i)\)
আবার,
\(\Rightarrow BC=AB\)
\(\Rightarrow BC^{2}=AB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-3)^{2}+(y-6)^{2}=(3-3)^{2}+(4-6)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.3+3^{2}+y^{2}-2.y.6+6^{2}=0^{2}+(-2)^{2}\)
\(\Rightarrow x^{2}-6x+9+y^{2}-12y+36=0+4\)
\(\Rightarrow x^{2}+y^{2}-6x-12y+45-4=0\)
\(\Rightarrow x^{2}+y^{2}-6x-12y+41=0 .......(ii)\)
এখন, \((i)-(ii)\) এর সাহায্যে পাই,
\(x^{2}+y^{2}-6x-8y+21-x^{2}-y^{2}+6x+12y-41=0\)
\(\Rightarrow -8y+21+12y-41=0\)
\(\Rightarrow 4y-20=0\)
\(\Rightarrow 4y=20\)
\(\therefore y=5\) ➜Note উভয় পার্শে \(4\) ভাগ করে।
\(y\) এর মান \((i)\)- এ বসিয়ে পাই,
\(x^{2}+5^{2}-6x-8\times 5+21=0\)
\(\Rightarrow x^{2}+25-6x-40+21=0\)
\(\Rightarrow x^{2}-6x+6=0\)
\(\Rightarrow x=\frac{-(-6)\pm\sqrt{(-6)^{2}-4.1.6}}{2.1}\)
\(\Rightarrow x=\frac{6\pm\sqrt{36-24}}{2}\)
\(\Rightarrow x=\frac{6\pm\sqrt{12}}{2}\)
\(\Rightarrow x=\frac{6\pm\sqrt{3\times 2^{2}}}{2}\)
\(\Rightarrow x=\frac{6\pm 2\sqrt{3}}{2}\)
\(\Rightarrow x=\frac{2(3\pm \sqrt{3})}{2}\)
\(\therefore x=3\pm \sqrt{3}\)
শর্তমতে, \(x=3+\sqrt{3}\) ➜Note \(\because C\) বিন্দুটি \(AB\) রেখার সাপেক্ষে মূলবিন্দুর বিপরীত পাশে অবস্থিত।
\(\therefore C\) বিন্দুর স্থানাঙ্ক \((3+\sqrt{3}, 5)\).

দেওয়া আছে, \(A(a+1, a)\), \(B(\sqrt{3}+1, 3\sqrt{3})\) এবং \(C(3\sqrt{3}+1, \sqrt{3})\)
এখন, \(AB=\sqrt{(a+1-\sqrt{3}-1)^{2}+(a-3\sqrt{3})^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(a-\sqrt{3})^{2}+(a-3\sqrt{3})^{2}}\)
\(\Rightarrow AB=\sqrt{a^{2}-2.a.\sqrt{3}+(\sqrt{3})^{2}+a^{2}-2.a.3\sqrt{3}+(3\sqrt{3})^{2}}\)
\(\Rightarrow AB=\sqrt{a^{2}-2a\sqrt{3}+3+a^{2}-6a\sqrt{3}+27}\)
\(\therefore AB=\sqrt{2a^{2}-8a\sqrt{3}+30}\)
আবার,
\(AC=\sqrt{(a+1-3\sqrt{3}-1)^{2}+(a-\sqrt{3})^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(a-3\sqrt{3})^{2}+(a-\sqrt{3})^{2}}\)
\(\Rightarrow AC=\sqrt{a^{2}-2.a.3\sqrt{3}+(3\sqrt{3})^{2}+a^{2}-2.a.\sqrt{3}+(\sqrt{3})^{2}}\)
\(\Rightarrow AC=\sqrt{a^{2}-6a\sqrt{3}+27+a^{2}-2a\sqrt{3}+3}\)
\(\therefore AC=\sqrt{2a^{2}-8a\sqrt{3}+30}\)
\(a\) এর যে কোন বাস্তব মানের জন্য \(AB=AC=\sqrt{2a^{2}-8a\sqrt{3}+30}\)
[ দেখানো হল।]
আবার,
\(BC=\sqrt{(\sqrt{3}+1-3\sqrt{3}-1)^{2}+(3\sqrt{3}-\sqrt{3})^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-2\sqrt{3})^{2}+(2\sqrt{3})^{2}}\)
\(\Rightarrow BC=\sqrt{12+12}\)
\(\therefore BC=\sqrt{24}\)
শর্তমতে, \(AB=AC=BC\)
\(\Rightarrow AB=BC\)
\(\Rightarrow AB^{2}=BC^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (\sqrt{2a^{2}-8a\sqrt{3}+30})^{2}=(\sqrt{24})^{2}\)
\(\Rightarrow 2a^{2}-8a\sqrt{3}+30=24\)
\(\Rightarrow 2a^{2}-8a\sqrt{3}+30-24=0\)
\(\Rightarrow 2a^{2}-8a\sqrt{3}+6=0\)
\(\Rightarrow a^{2}-4a\sqrt{3}+3=0\)
\(\Rightarrow a=\frac{-(-4\sqrt{3})\pm\sqrt{(-4\sqrt{3})^{2}-4.1.3}}{2.1}\)
\(\Rightarrow a=\frac{4\sqrt{3}\pm\sqrt{16\times 3-12}}{2}\)
\(\Rightarrow a=\frac{4\sqrt{3}\pm\sqrt{48-12}}{2}\)
\(\Rightarrow a=\frac{4\sqrt{3}\pm\sqrt{36}}{2}\)
\(\Rightarrow a=\frac{4\sqrt{3}\pm 6}{2}\)
\(\Rightarrow a=\frac{2(2\sqrt{3}\pm 3)}{2}\)
\(\Rightarrow a=2\sqrt{3}\pm 3\)
\(\therefore a=2\sqrt{3}\pm 3\).

মনে করি, \(A(3, 8)\),\(B(8, 3)\) এবং \(C(-2, 3)\).
এখন, \(AB=\sqrt{(3-8)^{2}+(8-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-5)^{2}+5^{2}}\)
\(\Rightarrow AB=\sqrt{25+25}\)
\(\Rightarrow AB=\sqrt{50}\)
\(\Rightarrow AB=\sqrt{2\times 5^{2}}\)
\(\therefore AB=5\sqrt{2}\)
আবার,
\(BC=\sqrt{(8+2)^{2}+(3-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{10^{2}+0^{2}}\)
\(\Rightarrow BC=\sqrt{100+0}\)
\(\Rightarrow BC=\sqrt{100}\)
\(\Rightarrow BC=\sqrt{10^{2}}\)
\(\therefore BC=10\)
আবার,
\(AC=\sqrt{(3+2)^{2}+(8-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{5^{2}+5^{2}}\)
\(\Rightarrow AC=\sqrt{25+25}\)
\(\Rightarrow AC=\sqrt{50}\)
\(\Rightarrow AC=\sqrt{2\times 5^{2}}\)
\(\therefore AC=5\sqrt{2}\)
\(\because AB=AC\neq BC\)
\(\therefore \) বিন্দু তিনটি একটি সমদ্বিবাহু ত্রিভুজের শীর্ষবিন্দু।

মনে করি, \(A(a, a)\),\(B(-a, -a)\) এবং \(C(-a\sqrt{3}, a\sqrt{3})\).
এখন, \(AB=\sqrt{(a+a)^{2}+(a+a)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(2a)^{2}+(2a)^{2}}\)
\(\Rightarrow AB=\sqrt{2\times (2a)^{2}}\)
\(\therefore AB=2a\sqrt{2}\)
আবার,
\(BC=\sqrt{(-a+a\sqrt{3})^{2}+(-a-a\sqrt{3})^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(a\sqrt{3}-a)^{2}+(a\sqrt{3}+a)^{2}}\)
\(\Rightarrow BC=\sqrt{2(a\sqrt{3})^{2}+2a^{2}}\)
\(\Rightarrow BC=\sqrt{6a^{2}+2a^{2}}\)
\(\Rightarrow BC=\sqrt{8a^{2}}\)
\(\Rightarrow BC=\sqrt{2\times (2a)^{2}}\)
\(\therefore BC=2a\sqrt{2}\)
আবার,
\(AC=\sqrt{(a+a\sqrt{3})^{2}+(a-a\sqrt{3})^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{2a^{2}+2(a\sqrt{3})^{2}}\)
\(\Rightarrow AC=\sqrt{2a^{2}+6a^{2}}\)
\(\Rightarrow AC=\sqrt{8a^{2}}\)
\(\Rightarrow AC=\sqrt{2\times (2a)^{2}}\)
\(\therefore AC=2a\sqrt{2}\)
\(\because AB=BC=AC\)
\(\therefore \) বিন্দু তিনটি একটি সমবাহু ত্রিভুজের শীর্ষবিন্দু।

মনে করি, \(A(12, 8)\),\(B(-2, 6)\) এবং \(C(6, 0)\).
এখন, \(AB=\sqrt{(12+2)^{2}+(8-6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{14^{2}+2^{2}}\)
\(\Rightarrow AB=\sqrt{196+4}\)
\(\Rightarrow AB=\sqrt{2\times 100}\)
\(\Rightarrow AB=\sqrt{2\times 10^{2}}\)
\(\therefore AB=10\sqrt{2}\)
আবার,
\(BC=\sqrt{(-2-6)^{2}+(6-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{8^{2}+6^{2}}\)
\(\Rightarrow BC=\sqrt{64+36}\)
\(\Rightarrow BC=\sqrt{100}\)
\(\Rightarrow BC=\sqrt{10^{2}}\)
\(\therefore BC=10\)
আবার,
\(AC=\sqrt{(12-6)^{2}+(8-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{6^{2}+8^{2}}\)
\(\Rightarrow AC=\sqrt{36+64}\)
\(\Rightarrow AC=\sqrt{100}\)
\(\Rightarrow AC=\sqrt{10^{2}}\)
\(\therefore AC=10\)
এখন,
\(BC^{2}+AC^{2}=10^{2}+10^{2}=100+100=200=(\sqrt{200})^{2}=(\sqrt{2\times 10^{2}})=(10\sqrt{2})^{2}=AB^{2}\)
\(\therefore BC^{2}+AC^{2}=AB^{2}\) ➜Note পিথাগোরাসের উপপাদ্য।
\(\therefore \) বিন্দু তিনটি একটি সমকোণী ত্রিভুজের শীর্ষবিন্দু।

মনে করি, \(A(4, 4)\),\(B(5, 2)\) এবং \(C(1, 0)\).
এখন, \(AB=\sqrt{(4-5)^{2}+(4-2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-1)^{2}+2^{2}}\)
\(\Rightarrow AB=\sqrt{1+4}\)
\(\Rightarrow AB=\sqrt{5}\)
\(\therefore AB=\sqrt{5}\)
আবার,
\(BC=\sqrt{(5-1)^{2}+(2-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{4^{2}+2^{2}}\)
\(\Rightarrow BC=\sqrt{16+4}\)
\(\Rightarrow BC=\sqrt{20}\)
\(\Rightarrow BC=\sqrt{5\times 2^{2}}\)
\(\therefore BC=2\sqrt{5}\)
আবার,
\(AC=\sqrt{(4-1)^{2}+(4-0)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{3^{2}+4^{2}}\)
\(\Rightarrow AC=\sqrt{9+16}\)
\(\Rightarrow AC=\sqrt{25}\)
\(\Rightarrow AC=\sqrt{5^{2}}\)
\(\therefore AC=5\)
এখন,
\(AB^{2}+BC^{2}=(\sqrt{5})^{2}+(2\sqrt{5})^{2}=5+20=25=5^{2}=AC^{2}\)
\(\therefore AB^{2}+BC^{2}=AC^{2}\) ➜Note পিথাগোরাসের উপপাদ্য।
\(\therefore \) বিন্দু তিনটি একটি সমকোণী ত্রিভুজের শীর্ষবিন্দু।
আবার,
\(\triangle ABC=\frac{1}{2}\times AB\times BC\)
\(\Rightarrow \triangle ABC=\frac{1}{2}\times \sqrt{5}\times 2\sqrt{5}\)
\(\Rightarrow \triangle ABC=5\) বর্গ একক।

দেওয়া আছে, \(A(3, 4)\),\(B(-4, 3)\) এবং \(C(4, -3)\).
এখন, \(AB=\sqrt{(3+4)^{2}+(4-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{7^{2}+1^{2}}\)
\(\Rightarrow AB=\sqrt{49+1}\)
\(\Rightarrow AB=\sqrt{50}\)
\(\Rightarrow AB=\sqrt{2\times 5^{2}}\)
\(\therefore AB=5\sqrt{2}\)
আবার,
\(BC=\sqrt{(-4-4)^{2}+(3+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-8)^{2}+6^{2}}\)
\(\Rightarrow BC=\sqrt{64+36}\)
\(\Rightarrow BC=\sqrt{100}\)
\(\Rightarrow BC=\sqrt{10^{2}}\)
\(\therefore BC=10\)
আবার,
\(AC=\sqrt{(3-4)^{2}+(4+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-1)^{2}+7^{2}}\)
\(\Rightarrow AC=\sqrt{1+49}\)
\(\Rightarrow AC=\sqrt{50}\)
\(\Rightarrow AC=\sqrt{2\times 5^{2}}\)
\(\therefore AC=5\sqrt{2}\)
এখানে,
\(\because AB=AC\neq BC\)
\(\therefore \triangle ABC\) সমদ্বিবাহু।
আবার,
\(AB^{2}+AC^{2}=(5\sqrt{2})^{2}+(5\sqrt{2})^{2}=50+50=100=10^{2}=BC^{2}\)
\(\therefore AB^{2}+AC^{2}=BC^{2}\) ➜Note পিথাগোরাসের উপপাদ্য।
\(\therefore \) বিন্দু তিনটি একটি সমদ্বিবাহু সমকোণী ত্রিভুজের শীর্ষবিন্দু।
আবার,
\(\triangle ABC=\frac{1}{2}\times AB\times AC\)
\(\Rightarrow \triangle ABC=\frac{1}{2}\times 5\sqrt{2}\times 5\sqrt{2}\)
\(\Rightarrow \triangle ABC=\frac{1}{2}\times 25\times 2\)
\(\Rightarrow \triangle ABC=25\) বর্গ একক।

মনে করি, \(A(1, 1)\), \(B(-4, 13)\),\(C(8, 8)\) এবং \(D(13, -4)\).
এখন, \(AB=\sqrt{(1+4)^{2}+(1-13)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{5^{2}+(-12)^{2}}\)
\(\Rightarrow AB=\sqrt{25+144}\)
\(\Rightarrow AB=\sqrt{169}\)
\(\Rightarrow AB=\sqrt{13^{2}}\)
\(\therefore AB=13\)
আবার,
\(BC=\sqrt{(-4-8)^{2}+(13-8)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-12)^{2}+5^{2}}\)
\(\Rightarrow BC=\sqrt{144+25}\)
\(\Rightarrow BC=\sqrt{169}\)
\(\Rightarrow BC=\sqrt{13^{2}}\)
\(\therefore BC=13\)
আবার,
\(CD=\sqrt{(8-13)^{2}+(8+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow CD=\sqrt{(-5)^{2}+12^{2}}\)
\(\Rightarrow CD=\sqrt{25+144}\)
\(\Rightarrow CD=\sqrt{169}\)
\(\Rightarrow CD=\sqrt{13^{2}}\)
\(\therefore CD=13\)
আবার,
\(AD=\sqrt{(1-13)^{2}+(1+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AD=\sqrt{(-12)^{2}+5^{2}}\)
\(\Rightarrow AD=\sqrt{144+25}\)
\(\Rightarrow AD=\sqrt{169}\)
\(\Rightarrow AD=\sqrt{13^{2}}\)
\(\therefore AD=13\)
\(\because AB=BC=CD=AD \)
\(\therefore ABCD \) চতুর্ভুজটি একটি বর্গক্ষেত্র অথবা, রম্বস হতে পারে।
আবার,
\(AC=\sqrt{(1-8)^{2}+(1-8)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-7)^{2}+(-7)^{2}}\)
\(\Rightarrow AC=\sqrt{49+49}\)
\(\Rightarrow AC=\sqrt{2\times 49}\)
\(\Rightarrow AC=\sqrt{2\times 7^{2}}\)
\(\therefore AC=7\sqrt{2}\)
আবার,
\(BD=\sqrt{(-4-13)^{2}+(13+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BD=\sqrt{(-17)^{2}+17^{2}}\)
\(\Rightarrow BD=\sqrt{25+144}\)
\(\Rightarrow BD=\sqrt{289+289}\)
\(\Rightarrow BD=\sqrt{2\times 289}\)
\(\Rightarrow BD=\sqrt{2\times 17^{2}}\)
\(\therefore BD=17\sqrt{2}\)
\(\because AB=BC=CD=AD \) এবং \(AC \neq BD \).
\(\therefore ABCD \) চতুর্ভুজটি একটি রম্বস।

দেওয়া আছে, \(A(a, b)\), \(B(a+\alpha, b+\beta)\),\(C(a+\alpha+p, b+\beta+q)\) এবং \(D(a+p, b+Q)\).
এখন, \(AB=\sqrt{(a-a-\alpha)^{2}+(b-b-\beta)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-\alpha)^{2}+(-\beta)^{2}}\)
\(\therefore AB=\sqrt{\alpha^{2}+\beta^{2}}\)
আবার,
\(BC=\sqrt{(a+\alpha-a-\alpha-p)^{2}+(b+\beta-b-\beta-q)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-p)^{2}+(-q)^{2}}\)
\(\therefore BC=\sqrt{p^{2}+q^{2}}\)
আবার,
\(CD=\sqrt{(a+\alpha+p-a-p)^{2}+(b+\beta+q-b-q)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\therefore CD=\sqrt{\alpha^{2}+\beta^{2}}\)
আবার,
\(AD=\sqrt{(a-a-p)^{2}+(b-b-q)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AD=\sqrt{(-p)^{2}+(-q)^{2}}\)
\(\therefore AD=\sqrt{p^{2}+q^{2}}\)
\(\because AB=CD, BC=AD \)
\(\therefore ABCD \) চতুর্ভুজটি একটি আয়তক্ষেত্র অথবা, সামান্তরিক হতে পারে।
আবার,
\(AC=\sqrt{(a-a-\alpha-p)^{2}+(b-b-\beta-q)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-\alpha-p)^{2}+(-\beta-q)^{2}}\)
\(\therefore AC=\sqrt{(\alpha+p)^{2}+(\beta+q)^{2}}\)
আবার,
\(BD=\sqrt{(a+\alpha-a-p)^{2}+(b+\beta-b-q)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\therefore BD=\sqrt{(\alpha-p)^{2}+(\beta-q)^{2}}\)
\(\because AB=CD, BC=AD \) এবং \(AC \neq BD \).
\(\therefore ABCD \) চতুর্ভুজটি একটি সামান্তরিক।
\((a)\) চতুর্ভুজটি একটি আয়তক্ষেত্র হবে যদি, \(AC=BD\) হয়।
সে ক্ষেত্রে \(\sqrt{(\alpha+p)^{2}+(\beta+q)^{2}}=\sqrt{(\alpha-p)^{2}+(\beta-q)^{2}}\)
\(\Rightarrow (\alpha+p)^{2}+(\beta+q)^{2}=(\alpha-p)^{2}+(\beta-q)^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (\alpha+p)^{2}-(\alpha-p)^{2}=(\beta-q)^{2}-(\beta+q)^{2}\)
\(\Rightarrow 4\alpha p=-4\beta q\) ➜Note\(\because (a+b)^{2}-(a-b)^{2}=4ab\)
\(\Rightarrow p\alpha=-q\beta \) ➜Note উভয় পার্শে \(4\) ভাগ করে।
\(\Rightarrow p\alpha+q\beta=0 \)
\(\therefore p\alpha+q\beta=0 \) হলে, চতুর্ভুজটি একটি আয়তক্ষেত্র হবে।
\((b)\) চতুর্ভুজটি একটি রম্বস হবে যদি, \(AB=BC\) অথবা, \(CD=AD \) হয়।
সে ক্ষেত্রে \(\sqrt{\alpha^{2}+\beta^{2}}=\sqrt{p^{2}+q^{2}}\)
\(\Rightarrow \alpha^{2}+\beta^{2}=p^{2}+q^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\therefore \alpha^{2}+\beta^{2}=p^{2}+q^{2}\) হলে, চতুর্ভুজটি একটি রম্বস হবে।

মনে করি, \(A(1, 3)\), \(B(5, 0)\), \(C(2, -4)\) এবং \(D(-2, -1)\).
এখন, \(AB=\sqrt{(1-5)^{2}+(3-0)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-4)^{2}+3^{2}}\)
\(\Rightarrow AB=\sqrt{16+9}\)
\(\Rightarrow AB=\sqrt{25}\)
\(\Rightarrow AB=\sqrt{5^{2}}\)
\(\therefore AB=5\)
আবার,
\(BC=\sqrt{(5-2)^{2}+(0+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{3^{2}+4^{2}}\)
\(\Rightarrow BC=\sqrt{9+16}\)
\(\Rightarrow BC=\sqrt{25}\)
\(\Rightarrow BC=\sqrt{5^{2}}\)
\(\therefore BC=5\)
আবার,
\(CD=\sqrt{(2+2)^{2}+(-4+1)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow CD=\sqrt{4^{2}+12^{2}}\)
\(\Rightarrow CD=\sqrt{16+9}\)
\(\Rightarrow CD=\sqrt{25}\)
\(\Rightarrow CD=\sqrt{5^{2}}\)
\(\therefore CD=5\)
আবার,
\(AD=\sqrt{(1+2)^{2}+(3+1)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AD=\sqrt{3^{2}+4^{2}}\)
\(\Rightarrow AD=\sqrt{9+16}\)
\(\Rightarrow AD=\sqrt{25}\)
\(\Rightarrow AD=\sqrt{5^{2}}\)
\(\therefore AD=5\)
\(\because AB=BC=CD=AD \)
\(\therefore ABCD \) চতুর্ভুজটি একটি বর্গক্ষেত্র অথবা, রম্বস হতে পারে।
আবার,
\(AC=\sqrt{(1-2)^{2}+(3+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-1)^{2}+7^{2}}\)
\(\Rightarrow AC=\sqrt{1+49}\)
\(\Rightarrow AC=\sqrt{50}\)
\(\Rightarrow AC=\sqrt{2\times 25}\)
\(\Rightarrow AC=\sqrt{2\times 5^{2}}\)
\(\therefore AC=5\sqrt{2}\)
আবার,
\(BD=\sqrt{(5+2)^{2}+(0+1)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BD=\sqrt{7^{2}+1^{2}}\)
\(\Rightarrow BD=\sqrt{49+1}\)
\(\Rightarrow BD=\sqrt{50}\)
\(\Rightarrow BD=\sqrt{2\times 25}\)
\(\Rightarrow BD=\sqrt{2\times 5^{2}}\)
\(\therefore BD=5\sqrt{2}\)
\(\because AB=BC=CD=AD \) এবং \(AC=BD \).
\(\therefore ABCD \) চতুর্ভুজটি একটি বর্গক্ষেত্র।

মনে করি, \(ABCD\) বর্গক্ষেত্রের \(AC\) কর্ণের প্রান্তবিন্দুর স্থানাঙ্ক যথক্রমে \(A(6, 3)\) ও \(C(-2, -3)\)।
এবং \(BD\) কর্ণের প্রান্তবিন্দুর স্থানাঙ্ক যথক্রমে \(B(x_{1}, y_{1})\) ও \(D(x_{2}, y_{2})\)
এখন,
\(AC=\sqrt{(6+2)^{2}+(3+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{8^{2}+6^{2}}\)
\(\Rightarrow AC=\sqrt{64+36}\)
\(\Rightarrow AC=\sqrt{100}\)
\(\Rightarrow AC=\sqrt{10^{2}}\)
\(\Rightarrow AC=10\)
আবার,
\(AB=BC=CD=AD=p\) হলে,
\(\Rightarrow AB^{2}+BC^{2}=AC^{2}\)
\(\Rightarrow p^{2}+p^{2}=10^{2}\)
\(\Rightarrow 2p^{2}=100\)
\(\Rightarrow p^{2}=\frac{100}{2}\)
\(\Rightarrow p^{2}=50\)
\(\Rightarrow p=\sqrt{50}\)
\(\therefore AB=BC=CD=AD=\sqrt{50}\)
এখন,
\(\therefore \Box ABCD\) \(= AB\times BC\) \(= \sqrt{50}\times \sqrt{50}\) \(= 50\) বর্গ একক।
আবার,
\(BA=BC=\sqrt{50}\)
\(\Rightarrow BA^{2}=BC^{2}=(\sqrt{50})^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x_{1}-6)^{2}+(y_{1}-3)^{2}=(x_{1}+2)^{2}+(y_{1}+3)^{2}=50\)
\(\Rightarrow (x_{1}-6)^{2}+(y_{1}-3)^{2}=(x_{1}+2)^{2}+(y_{1}+3)^{2}\)
\(\Rightarrow (x_{1})^{2}-2.x_{1}.6+6^{2}+(y_{1})^{2}-2.y_{1}.3+3^{2}=(x_{1})^{2}+2.x_{1}.2+2^{2}+(y_{1})^{2}+2.y_{1}.3+3^{2}\)
\(\Rightarrow (x_{1})^{2}-12x_{1}+36+(y_{1})^{2}-6y_{1}+9-(x_{1})^{2}-4x_{1}-4-(y_{1})^{2}-6y_{1}-9=0\)
\(\Rightarrow -16x_{1}-12y_{1}+32=0\)
\(\Rightarrow 4x_{1}+3y_{1}-8=0\) ➜Note উভয় পার্শে \(-4\) ভাগ করে।
\(\Rightarrow 3y_{1}=8-4x_{1}\)
\(\therefore y_{1}=\frac{8-4x_{1}}{3} .....(i)\)
আবার,
\( (x_{1}+2)^{2}+(y_{1}+3)^{2}=50\)
\(\Rightarrow x_{1}^{2}+2.x_{1}.2+2^{2}+y_{1}^{2}+2.y_{1}.3+3^{2}=50\)
\(\Rightarrow x_{1}^{2}+y_{1}^{2}+4x_{1}+4+6y_{1}+9-50=0\)
\(\Rightarrow x_{1}^{2}+(\frac{8-4x_{1}}{3})^{2}+4x_{1}+6\times \frac{8-4x_{1}}{3}-37=0\) ➜Note \((i)\) এর সাহায্যে
\(\Rightarrow x_{1}^{2}+\frac{(8-4x_{1})^{2}}{9}+4x_{1}+2(8-4x_{1})-37=0\)
\(\Rightarrow 9x_{1}^{2}+(8-4x_{1})^{2}+36x_{1}+18(8-4x_{1})-333=0\) ➜Note উভয় পার্শে \(9\) গুণ করে।
\(\Rightarrow 9x_{1}^{2}+(8)^{2}-2.8.4x_{1}+(4x_{1})^{2}+36x_{1}+144-72x_{1}-333=0\)
\(\Rightarrow 9x_{1}^{2}+64-64x_{1}+16x_{1}^{2}+36x_{1}+144-72x_{1}-333=0\)
\(\Rightarrow 25x_{1}^{2}-100x_{1}-125=0\)
\(\Rightarrow x_{1}^{2}-4x_{1}-5=0\) ➜Note উভয় পার্শে \(25\) ভাগ করে।
\(\Rightarrow x_{1}^{2}-5x_{1}+x_{1}-5=0\)
\(\Rightarrow x_{1}( x_{1}-5)+1(x_{1}-5)=0\)
\(\Rightarrow (x_{1}-5)(x_{1}+1)=0\)
\(\Rightarrow (x_{1}-5)=0, (x_{1}+1)=0\)
\(\Rightarrow x_{1}=5, x_{1}=-1\)
\(\therefore x_{1}=5, x_{2}=-1\)
এখন, \((i)\) হতে,
যখন, \(x_{1}=5\) তখন, \(y_{1}=\frac{8-4.5}{3}=\frac{8-20}{3}=\frac{-12}{3}=-4\)
আবার,
যখন, \(x_{2}=-1\) তখন, \(y_{2}=\frac{8-4.(-1)}{3}=\frac{8+4}{3}=\frac{12}{3}=4\)
\(\therefore\) বর্গের অপর দুইটি শীর্ষবিন্দুর স্থানাঙ্ক \((5, -4)\), \((-1, 4)\)

মনে করি, \(A(-5, 1)\), \(B(3, -3)\), \(C(1, -7)\) এবং \(D(-7, -3)\).
এখন, \(AB=\sqrt{(-5-3)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-8)^{2}+4^{2}}\)
\(\Rightarrow AB=\sqrt{64+16}\)
\(\Rightarrow AB=\sqrt{80}\)
\(\Rightarrow AB=\sqrt{5\times 16}\)
\(\Rightarrow AB=\sqrt{5\times 4^{2}}\)
\(\therefore AB=4\sqrt{5}\)
আবার,
\(BC=\sqrt{(3-1)^{2}+(-3+7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{2^{2}+4^{2}}\)
\(\Rightarrow BC=\sqrt{4+16}\)
\(\Rightarrow BC=\sqrt{20}\)
\(\Rightarrow BC=\sqrt{5\times 2^{2}}\)
\(\therefore BC=2\sqrt{5}\)
আবার,
\(CD=\sqrt{(1+7)^{2}+(-7+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow CD=\sqrt{8^{2}+(-4)^{2}}\)
\(\Rightarrow CD=\sqrt{64+16}\)
\(\Rightarrow CD=\sqrt{80}\)
\(\Rightarrow CD=\sqrt{5\times 16}\)
\(\Rightarrow CD=\sqrt{5\times 4^{2}}\)
\(\therefore CD=4\sqrt{5}\)
আবার,
\(AD=\sqrt{(-5+7)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AD=\sqrt{(2^{2}+4^{2}}\)
\(\Rightarrow AD=\sqrt{4+16}\)
\(\Rightarrow AD=\sqrt{20}\)
\(\Rightarrow AD=\sqrt{5\times 2^{2}}\)
\(\therefore AD=2\sqrt{5}\)
\(\because AB=CD, AD=BC \)
\(\therefore ABCD \) চতুর্ভুজটি একটি আয়তক্ষেত্র অথবা, সামান্তরিক হতে পারে।
আবার,
\(AC=\sqrt{(-5-1)^{2}+(1+7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-6)^{2}+8^{2}}\)
\(\Rightarrow AC=\sqrt{36+64}\)
\(\Rightarrow AC=\sqrt{100}\)
\(\Rightarrow AC=\sqrt{10^{2}}\)
\(\therefore AC=10\)
আবার,
\(BD=\sqrt{(3+7)^{2}+(-3+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BD=\sqrt{10^{2}+0^{2}}\)
\(\Rightarrow BD=\sqrt{10^{2}}\)
\(\therefore BD=10\)
\(\because AB=CD, AD=BC \) এবং \(AC=BD \).
\(\therefore ABCD \) চতুর্ভুজটি একটি আয়তক্ষেত্র।
আবার,
\(\therefore \Box ABCD\) \(= AB\times BC\)
\(= 4\sqrt{5}\times 2\sqrt{5}\)
\(= 8\times 5=40\) বর্গ একক।

মনে করি, \(A(-2, -1)\), \(B(1, 0)\), \(C(4, 3)\) এবং \(D(1, 2)\).
এখন, \(AB=\sqrt{(-2-1)^{2}+(-1-0)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{(-3)^{2}+(-1)^{2}}\)
\(\Rightarrow AB=\sqrt{9+1}\)
\(\therefore AB=\sqrt{10}\)
আবার,
\(BC=\sqrt{(1-4)^{2}+(0-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BC=\sqrt{(-3)^{2}+(-3)^{2}}\)
\(\Rightarrow BC=\sqrt{9+9}\)
\(\Rightarrow BC=\sqrt{18}\)
\(\Rightarrow BC=\sqrt{2\times 3^{2}}\)
\(\therefore BC=3\sqrt{2}\)
আবার,
\(CD=\sqrt{(4-1)^{2}+(3-2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow CD=\sqrt{3^{2}+1^{2}}\)
\(\Rightarrow CD=\sqrt{9+1}\)
\(\therefore CD=\sqrt{10}\)
আবার,
\(AD=\sqrt{(-2-1)^{2}+(-1-2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AD=\sqrt{((-3)^{2}+(-3)^{2}}\)
\(\Rightarrow AD=\sqrt{9+9}\)
\(\Rightarrow AD=\sqrt{18}\)
\(\Rightarrow AD=\sqrt{2\times 3^{2}}\)
\(\therefore AD=3\sqrt{2}\)
\(\because AB=CD, AD=BC \)
\(\therefore ABCD \) চতুর্ভুজটি একটি আয়তক্ষেত্র অথবা, সামান্তরিক হতে পারে।
আবার,
\(AC=\sqrt{(-2-4)^{2}+(-1-3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AC=\sqrt{(-6)^{2}+(-4)^{2}}\)
\(\Rightarrow AC=\sqrt{36+16}\)
\(\Rightarrow AC=\sqrt{52}\)
\(\Rightarrow AC=\sqrt{13\times 2^{2}}\)
\(\therefore AC=2\sqrt{13}\)
আবার,
\(BD=\sqrt{(1-1)^{2}+(0-2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow BD=\sqrt{0^{2}+(-2)^{2}}\)
\(\Rightarrow BD=\sqrt{0+4}\)
\(\Rightarrow BD=\sqrt{2^{2}}\)
\(\therefore BD=2\)
\(\because AB=CD, AD=BC \) এবং \(AC \neq BD \).
\(\therefore ABCD \) চতুর্ভুজটি একটি সামান্তরিক।

মনে করি, \(A(5, 7)\), \(B(-1, -1)\), \(C(-2, 6)\) এবং কেন্দ্র \(P(x, y)\).
শর্তমতে, \(PA=PB=PC\)
\(\Rightarrow PA^{2}=PB^{2}=PC^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow PA^{2}=PB^{2}\)
\(\Rightarrow (x-5)^{2}+(y-7)^{2}=(x+1)^{2}+(y+1)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.5+5^{2}+y^{2}-2.y.7+7^{2}=x^{2}+2.x.1+1^{2}+y^{2}+2.y.1+1^{2}\)
\(\Rightarrow x^{2}-10x+25+y^{2}-14y+49-x^{2}-2x-1-y^{2}-2y-1=0\)
\(\Rightarrow -12x-16y+72=0\)
\(\Rightarrow 3x+4y-18=0........(i)\) ➜Note উভয় পার্শে \(-4\) ভাগ করে।
আবার,
\(\Rightarrow PA^{2}=PC^{2}\)
\(\Rightarrow (x-5)^{2}+(y-7)^{2}=(x+2)^{2}+(y-6)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.5+5^{2}+y^{2}-2.y.7+7^{2}=x^{2}+2.x.2+2^{2}+y^{2}-2.y.6+6^{2}\)
\(\Rightarrow x^{2}-10x+25+y^{2}-14y+49-x^{2}-4x-4-y^{2}+12y-36=0\)
\(\Rightarrow -14x-2y+34=0\)
\(\Rightarrow 7x+y-17=0 .....(ii)\) ➜Note উভয় পার্শে \(-2\) ভাগ করে।
এখন,
\((ii)\times 4-(i)\) এর সাহায্যে,
\(28x+4y-68-3x-4y+18=0\)
\(\Rightarrow 25x-50=0\)
\(\Rightarrow 25x=50\)
\(\Rightarrow x=\frac{50}{25}\)
\(\therefore x=2\)
আবার,
\((i)\) হতে \( 3\times 2+4y-18=0\) ➜Note \(\because x=2\)
\(\Rightarrow 6+4y-18=0\)
\(\Rightarrow 4y=18-6\)
\(\Rightarrow 4y=12\)
\(\Rightarrow y=\frac{12}{4}\)
\(\therefore y=3\)
\(\therefore \) কেন্দ্রের স্থানাঙ্ক \((2, 3)\)

মনে করি, বৃত্তের কেন্দ্র \(C(5, 3)\) এবং \(AB\) জ্যা-এর মধ্যবিন্দু \(D(3, 2)\)
শর্তমতে, বৃত্তের ব্যাসার্ধ \(CA=CB=5\)
\(C,D\) যোগ করি। তাহলে, \(CD \bot AB\) ➜Note বৃত্তের কেন্দ্র হতে জ্যা-এর মধ্যবিন্দুর সংযোগ সরলরেখা জ্যা-এর উপর লম্ব।
এখন, \(CD^{2}=(5-3)^{2}+(3-2)^{2}\) \(=2^{2}+1^{2}\) \(=4+1\) \(=5\)
\(CAD\) সমকোণী ত্রিভুজ হতে, \(CA^{2}=AD^{2}+CD^{2}\) \(\Rightarrow 5^{2}=AD^{2}+5\) ➜Note\(\because CA=5, CD^{2}=5\)
\(\Rightarrow 25=AD^{2}+5\)
\(\Rightarrow AD^{2}+5=25\)
\(\Rightarrow AD^{2}=25-5\)
\(\Rightarrow AD^{2}=20\)
\(\Rightarrow AD=\sqrt{20}\)
\(\Rightarrow AD=\sqrt{5\times 2^{2}}\)
\(\therefore AD=2\sqrt{5}\)
আবার, জ্যা-এর দৈর্ঘ্য \(AB=2\times AD\) \(=2\times 2\sqrt{5}\) \(=4\sqrt{5}\) ➜Note\(\because AD=2\sqrt{5}\)
\(\therefore \) জ্যা-এর দৈর্ঘ্য \(=4\sqrt{5}\).

মনে করি, \(O(0, 0)\), \(A(0, 8)\), \(B(4, 0)\) এবং পরিকেন্দ্র \(C(x, y)\).
শর্তমতে, \(CO=CA=CB\)
\(\Rightarrow CO^{2}=CA^{2}=CB^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow CO^{2}=CA^{2}\)
\(\Rightarrow (x-0)^{2}+(y-0)^{2}=(x-0)^{2}+(y-8)^{2}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}+y^{2}=x^{2}+y^{2}-2.y.8+8^{2}\)
\(\Rightarrow x^{2}+y^{2}-x^{2}-y^{2}+16y-64=0\)
\(\Rightarrow 16y-64=0\)
\(\Rightarrow 16y=64\)
\(\Rightarrow y=4\) ➜Note উভয় পার্শে \(16\) ভাগ করে।
আবার,
\(\Rightarrow CO^{2}=CB^{2}\)
\(\Rightarrow (x-0)^{2}+(y-0)^{2}=(x-4)^{2}+(y-0)^{2}\)
\(\Rightarrow x^{2}+y^{2}=x^{2}-2.x.4+4^{2}+y^{2}\)
\(\Rightarrow x^{2}+y^{2}-x^{2}+8x-16-y^{2}=0\)
\(\Rightarrow 8x-16=0\)
\(\Rightarrow 8x=16\)
\(\Rightarrow x=2\) ➜Note উভয় পার্শে \(8\) ভাগ করে।
\(\therefore \) পরিকেন্দ্রের স্থানাঙ্ক \((2, 4)\)

মনে করি, \(A(-1, 5)\), \(B(6, -2)\), \(C(7, -1)\), \(D(0,-2)\) এবং কেন্দ্র \(P(3, 2)\).
এখন,
\(PA=\sqrt{(3+1)^{2}+(2-5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow PA=\sqrt{4^{2}+(-3)^{2}}\)
\(\Rightarrow PA=\sqrt{16+9}\)
\(\Rightarrow PA=\sqrt{25}\)
\(\Rightarrow PA=\sqrt{5^{2}}\)
\(\Rightarrow PA=5\)
আবার,
\(PB=\sqrt{(3-6)^{2}+(2+2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow PB=\sqrt{(-3)^{2}+4^{2}}\)
\(\Rightarrow PB=\sqrt{9+16}\)
\(\Rightarrow PB=\sqrt{25}\)
\(\Rightarrow PB=\sqrt{5^{2}}\)
\(\Rightarrow PB=5\)
আবার,
\(PC=\sqrt{(3-7)^{2}+(2+1)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow PC=\sqrt{(-4)^{2}+3^{2}}\)
\(\Rightarrow PC=\sqrt{16+9}\)
\(\Rightarrow PC=\sqrt{25}\)
\(\Rightarrow PC=\sqrt{5^{2}}\)
\(\Rightarrow PC=5\)
আবার,
\(PD=\sqrt{(3-0)^{2}+(2+2)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow PD=\sqrt{3^{2}+4^{2}}\)
\(\Rightarrow PD=\sqrt{9+16}\)
\(\Rightarrow PD=\sqrt{25}\)
\(\Rightarrow PD=\sqrt{5^{2}}\)
\(\Rightarrow PD=5\)
\(\because PA=PB=PC=PD=5=\) বৃত্তের ব্যাসার্ধ
[ প্রমাণিত ]

মনে করি, ব্যাসের প্রান্তবিন্দুদ্বয় \(A(5, 2)\) এবং \(B(-3, -4)\)।
\(\therefore \) ব্যাসের দৈর্ঘ্য \(AB=\sqrt{(5+3)^{2}+(2+4)^{2}}\)➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(\Rightarrow AB=\sqrt{8^{2}+6^{2}}\)
\(\Rightarrow AB=\sqrt{64+36}\)
\(\Rightarrow AB=\sqrt{100}\)
\(\Rightarrow AB=\sqrt{10^{2}}\)
\(\Rightarrow AB=10\)
এখন,
ব্যাসার্ধ \(=\frac{AB}{2}\) \(=\frac{10}{2}\) \(=5\)
\(\therefore\) ব্যাসার্ধ \(=5\) একক।

\((a)\)
মূলবিন্দু হতে \(C(2, -1)\) বিন্দুর দূরত্ব \(OC=\sqrt{(0-2)^{2}+(0+1)^{2}}\)
\(\Rightarrow OC=\sqrt{(-2)^{2}+1^{2}}\)
\(\Rightarrow OC=\sqrt{4+1}\)
\(\therefore OC=\sqrt{5}\) একক।
আবার,
\(X\) অক্ষ হতে \(C(2, -1)\) বিন্দুর দূরত্ব \(=|-1|=1\) একক।
\(Y\) অক্ষ হতে \(C(2, -1)\) বিন্দুর দূরত্ব \(=|2|=2\) একক।
\((b)\)
মনে করি, বিন্দুরটির ভুজ \(x\) একক।
\(\therefore \) বিন্দুটির কটি \(2x\) একক।
\(\therefore \) বিন্দুটি \(P(x, 2x)\).
এবং \(A(4, 3)\).
শর্তমতে, \(PA=\sqrt{10}\)
\(\Rightarrow PA^{2}=(\sqrt{10})^{2}\) ➜Note উভয় পার্শে বর্গ করে।
\(\Rightarrow (x-4)^{2}+(2x-3)^{2}=10\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(\Rightarrow x^{2}-2.x.4+4^{2}+(2x)^{2}-2.2x.3+3^{2}=10\)
\(\Rightarrow x^{2}-8x+16+4x^{2}-12x+9=10\)
\(\Rightarrow 5x^{2}-20x+25-10=0\)
\(\Rightarrow 5x^{2}-20x+15=0\)
\(\Rightarrow x^{2}-4x+3=0\)
\(\Rightarrow x^{2}-3x-x+3=0\)
\(\Rightarrow x(x-3)-1(x-3)=0\)
\(\Rightarrow (x-3)(x-1)=0\)
\(\Rightarrow (x-3)=0, (x-1)=0\)
\(\Rightarrow x=3, x=1\)
\(\therefore \) বিন্দুটি \(P(3, 6)\), \(P(1, 2)\).
\((c)\)
দেওয়া আছে, বৃত্তের কেন্দ্র \(B(11, 2)\) এবং \(PQ\) জ্যা-এর মধ্যবিন্দু \(C(2, -1)\)
শর্তমতে, বৃত্তের ব্যাসার্ধ \(PB=QB=10\)
\(C,B\) যোগ করি। তাহলে, \(CB \bot PC\) ➜Note বৃত্তের কেন্দ্র হতে জ্যা-এর মধ্যবিন্দুর সংযোগ সরলরেখা জ্যা-এর উপর লম্ব।
এখন, \(CB^{2}=(2-11)^{2}+(-1-2)^{2}\)
\(=(-9)^{2}+(-3)^{2}\)
\(=81+9\)
\(=90\)
\(PCB\) সমকোণী ত্রিভুজ হতে, \(PB^{2}=PC^{2}+CB^{2}\)
\(\Rightarrow 10^{2}=PC^{2}+90\) ➜Note \(\because PB=5, CB^{2}=90\)
\(\Rightarrow 100=PC^{2}+90\)
\(\Rightarrow PC^{2}+90=100\)
\(\Rightarrow PC^{2}=100-90\)
\(\Rightarrow PC^{2}=10\)
\(\therefore PC=\sqrt{10}\)
আবার, জ্যা-এর দৈর্ঘ্য \(PQ=2\times PC\)
\(=2\times \sqrt{10}\)
\(=2\sqrt{10}\) ➜Note \(\because PC=\sqrt{10}\)
\(\therefore \) জ্যা-এর দৈর্ঘ্য \(=2\sqrt{10}\) একক।

\((a)\)
মনে করি, \(P(x_{1}, y_{1})\) এবং \(Q(x_{2}, y_{2})\) একই সমতলে যে কোন দুইটি বিন্দু।
\(\therefore \) \(PQ\) দূরত্ব \(=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\).
\((b)\)
\(ABCD\) রম্বসের কর্ণদ্বয় হবে যথাক্রমে \(AC\) এবং \(BD\)।
\(\therefore \) কর্ণ \(AC=\sqrt{(1-8)^{2}+(1-8)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(=\sqrt{(-7)^{2}+(-7)^{2}}\)
\(=\sqrt{49+49}\)
\(=\sqrt{2\times 49}\)
\(=\sqrt{2\times 7^{2}}\)
\(=7\sqrt{2}\)
এবং কর্ণ \(BD=\sqrt{(-4-13)^{2}+(13+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-17)^{2}+17^{2}}\)
\(=\sqrt{289+289}\)
\(=\sqrt{2\times 289}\)
\(=\sqrt{2\times 17^{2}}\)
\(=17\sqrt{2}\)
\(\therefore \) রম্বসের কর্ণদ্বয় যথাক্রমে \(AC=7\sqrt{2}\) এবং \(BD=17\sqrt{2}\)।
\((c)\)
দেওয়া আছে, \(A(1, 1)\), \(B(-4, 13)\), \(C(8, 8)\) এবং \(D(13, -4)\).
এখন,
\(AB=\sqrt{(1+4)^{2}+(1-13)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB^2=(x_1-x_2)^2+(y_1-y_2)^2\)
\(=\sqrt{5^{2}+(-12)^{2}}\)
\(=\sqrt{25+144}\)
\(=\sqrt{169}\)
\(=\sqrt{13^{2}}\)
\(=13\)
\(BC=\sqrt{(-4-8)^{2}+(13-8)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-12)^{2}+5^{2}}\)
\(=\sqrt{144+25}\)
\(=\sqrt{169}\)
\(=\sqrt{13^{2}}\)
\(=13\)
\(CD=\sqrt{(8-13)^{2}+(8+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-5)^{2}+12^{2}}\)
\(=\sqrt{25+144}\)
\(=\sqrt{169}\)
\(=\sqrt{13^{2}}\)
\(=13\)
\(AD=\sqrt{(1-13)^{2}+(1+4)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-12)^{2}+5^{2}}\)
\(=\sqrt{144+25}\)
\(=\sqrt{169}\)
\(=\sqrt{13^{2}}\)
\(=13\)
আবার,
\((b)\) হতে প্রাপ্ত রম্বসের কর্ণদ্বয় যথাক্রমে \(AC=7\sqrt{2}\) এবং \(BD=17\sqrt{2}\)।
\(\because AB=BC=CD=AD\) এবং \(AC\neq BD\)
\(\therefore \Box ABCD\) একটি রম্বস।
আবার,
\(\therefore \Box ABCD=\frac{1}{2}\times AC\times BD\)
\(=\frac{1}{2}\times 7\sqrt{2}\times 17\sqrt{2}\)
\(=\frac{1}{2}\times 119\times (\sqrt{2})^{2}\)
\(=\frac{1}{2}\times 119\times 2\)
\(= 119\) বর্গ একক।

\((a)\)
স্থানাংকঃ বিন্দুর অবস্থান অঙ্কের সাহায্যে প্রকাশ করা হলে, ঐ অংকগুলিকে বিন্দুর স্থানাঙ্ক বলে।
যেমনঃ \(A(3, -7)\), \(B(3, 2)\)...........\(P(x, y)\). এখানে \(x\) এবং \(y\) যে কোন বাস্তব সংখ্যা। দ্বিমাত্রিক জ্যামিতিতে বিন্দুর দুই প্রকারের স্থানাংকের উল্লেখ আছে। যেমনঃ কার্তেসীয় স্থানাংক এবং পোলার স্থানাংক।
কার্তেসীয় স্থানাংকঃ

পোলার স্থানাংকঃ

\((b)\)
দেওয়া আছে,\(A(5, 6)\),\(B(-6, 5)\) এবং \(C(-5, -6)\)
এখন,
\(AB=\sqrt{(5+6)^{2}+(6-5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{11^{2}+1^{2}}\)
\(=\sqrt{121+1}\)
\(=\sqrt{122}\)
\(BC=\sqrt{(-6+5)^{2}+(5+6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-1)^{2}+11^{2}}\)
\(=\sqrt{1+121}\)
\(=\sqrt{122}\)
\(CA=\sqrt{(-5-5)^{2}+(-6-6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-10)^{2}+(-12)^{2}}\)
\(=\sqrt{100+144}\)
\(=\sqrt{244}\)
এখানে, \(AB^{2}+BC^{2}=(\sqrt{122})^{2}+(\sqrt{122})^{2}=122+122=244=(\sqrt{244})^{2}=CA^{2}\)
\(\because AB^{2}+BC^{2}=CA^{2}\) এবং \(AB=BC\)
\(\therefore \triangle ABC\) সমদ্বিবাহু সমকোণী।
\((c)\)
দেওয়া আছে,\(A(5, 6)\),\(B(-6, 5)\), \(C(-5, -6)\) এবং \(D(6, -5)\)
এখন,
\(AB=\sqrt{(5+6)^{2}+(6-5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{11^{2}+1^{2}}\)
\(=\sqrt{121+1}\)
\(=\sqrt{122}\)
\(BC=\sqrt{(-6+5)^{2}+(5+6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-1)^{2}+11^{2}}\)
\(=\sqrt{1+121}\)
\(=\sqrt{122}\)
\(CD=\sqrt{(-5-6)^{2}+(-6+5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-11)^{2}+(-1)^{2}}\)
\(=\sqrt{121+1}\)
\(=\sqrt{122}\)
\(DA=\sqrt{(6-5)^{2}+(-5-6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{1^{2}+(-11)^{2}}\)
\(=\sqrt{1+121}\)
\(=\sqrt{122}\)
\(AC=\sqrt{(5+5)^{2}+(6+6)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{10^{2}+12^{2}}\)
\(=\sqrt{100+144}\)
\(=\sqrt{244}\)
\(BD=\sqrt{(-6-6)^{2}+(5+5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-12)^{2}+10^{2}}\)
\(=\sqrt{144+100}\)
\(=\sqrt{244}\)
\(\because AB=BC=CD=DA\) এবং \(AC=BD\)
\(\therefore \Box ABCD\) একটি বর্গক্ষেত্র।
এবং \(\Box ABCD=AB\times BC\)
\(=\sqrt{122}\times \sqrt{122}\)
\(=(\sqrt{122})^{2}\)
\(=122\) বর্গ একক।

\((a)\)
\(r=\sqrt{x^{2}+y^{2}}\)
\(\theta=tan^{-1}\frac{y}{x}\)
এবং
\(x=r\cos\theta\)
\(y=r\sin\theta\)
\((b)\)
দেওয়া আছে, \(A(-5, 1)\), \(B(3, -3)\) এবং \(C(1, -7)\)
এখন,
\(AB=\sqrt{(-5-3)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-8)^{2}+4^{2}}\)
\(=\sqrt{64+16}\)
\(=\sqrt{80}\)
\(=\sqrt{5\times 16}\)
\(=\sqrt{5\times 4^{2}}\)
\(=4\sqrt{5}\)
\(BC=\sqrt{(3-1)^{2}+(-3+7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{2^{2}+4^{2}}\)
\(=\sqrt{4+16}\)
\(=\sqrt{20}\)
\(=\sqrt{5\times 4}\)
\(=\sqrt{5\times 2^{2}}\)
\(=2\sqrt{5}\)
\(CA=\sqrt{(1+5)^{2}+(-7-1)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{6^{2}+(-8)^{2}}\)
\(=\sqrt{36+64}\)
\(=\sqrt{100}\)
\(=\sqrt{10^{2}}\)
\(=10\)
এখানে, \(AB^{2}+BC^{2}=(4\sqrt{5})^{2}+(2\sqrt{5})^{2}=16\times 5+4\times 5=80+20=100=(\sqrt{10})^{2}=CA^{2}\)
\(\because AB^{2}+BC^{2}=CA^{2}\)
\(\therefore \triangle ABC\) সমকোণী।
অতিভুজ \(AC=10\) একক।
\(\triangle ABC=\frac{1}{2}\times 4\sqrt{5}\times 2\sqrt{5}\)
\(\Rightarrow \triangle ABC=\frac{1}{2}\times 8\times (\sqrt{5})^{2}\)
\(\Rightarrow \triangle ABC=4\times 5\)
\(\therefore \triangle ABC=20\) বর্গ একক।
\((c)\)
দেওয়া আছে, \(A(-5, 1)\), \(B(3, -3)\), \(C(1, -7)\) এবং \(D(-7, -3)\)
\(AB=\sqrt{(-5-3)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-8)^{2}+4^{2}}\)
\(=\sqrt{64+16}\)
\(=\sqrt{80}\)
\(=\sqrt{5\times 16}\)
\(=\sqrt{5\times 4^{2}}\)
\(=4\sqrt{5}\)
\(BC=\sqrt{(3-1)^{2}+(-3+7)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{2^{2}+4^{2}}\)
\(=\sqrt{4+16}\)
\(=\sqrt{20}\)
\(=\sqrt{5\times 4}\)
\(=\sqrt{5\times 2^{2}}\)
\(=2\sqrt{5}\)
\(CD=\sqrt{(1+7)^{2}+(-7+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{8^{2}+(-4)^{2}}\)
\(=\sqrt{64+16}\)
\(=\sqrt{80}\)
\(=\sqrt{5\times 16}\)
\(=\sqrt{5\times 4^{2}}\)
\(=4\sqrt{5}\)
\(AD=\sqrt{(-5+7)^{2}+(1+3)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{2^{2}+4^{2}}\)
\(=\sqrt{4+16}\)
\(=\sqrt{20}\)
\(=\sqrt{5\times 4}\)
\(=\sqrt{5\times 2^{2}}\)
\(=2\sqrt{5}\)
\(AC=\sqrt{(-7-1)^{2}+(1+5)^{2}}\) ➜Note \(A(x_1, y_1)\) এবং \(B(x_2, y_2)\), \(AB=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)
\(=\sqrt{(-8)^{2}+6^{2}}\)
\(=\sqrt{64+36}\)
\(=\sqrt{100}\)
\(=\sqrt{10^{2}}\)
\(=10\)
\(=\sqrt{10^{2}+0^{2}}\)
\(=\sqrt{100+0}\)
\(=\sqrt{100}\)
\(=\sqrt{10^{2}}\)
\(=10\)
\(\because AB=CD, BC=AD\) এবং \(AC=BD\)
\(\therefore \Box ABCD\) একটি আয়তক্ষেত্র।
\(\because \) আয়তক্ষেত্রের বিপরীত কোণগুলির যোগফল \(180^{o}\)
\(\therefore \Box ABCD\) আয়তক্ষেত্রটি বৃত্তস্থ।
\(\therefore \Box ABCD\) আয়তক্ষেত্রের পরিকেন্দ্রের স্থানাঙ্ক হবে \(AC\) বা \(BD\) কর্ণের মধ্যবিন্দু
\(\therefore \) পরিকেন্দ্রের স্থানাঙ্ক \((\frac{-5+1}{2}, \frac{1-7}{2})\)
\(\Rightarrow (\frac{-4}{2}, \frac{-6}{2})\)
\(\Rightarrow (-2, -3)\)
\(\therefore \) পরিকেন্দ্রের স্থানাঙ্ক \(P(-2, -3)\)