প্রমাণঃ
ধরি,
\(P(x_{1}, y_{1})\)
\(ax+by+c=0 ..............(1)\)
\((1)\) এর উপর যে কোনো বিন্দু \(Q(x_{2}, y_{2})\)
\((1)\) এর ঢাল \(m_{1}=-\frac{a}{b}\)
\(PQ\) এর ঢাল \(m_{2}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\)
শর্তমতে,
\((1)\) সরলরেখা এবং \(PQ\) রেখাংশ পরস্পর লম্ব।
\(\therefore m_{1}\times m_{2}=-1\)
\(\Rightarrow -\frac{a}{b}\times \frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-1\)
\(\Rightarrow \frac{ay_{1}-ay_{2}}{bx_{1}-bx_{2}}=1\)
\(\Rightarrow ay_{1}-ay_{2}=bx_{1}-bx_{2}\)
\(\Rightarrow ay_{1}-ay_{2}-bx_{1}+bx_{2}=0\)
\(\Rightarrow bx_{2}-bx_{1}+ay_{1}-ay_{2}=0 ...........(2)\)
আবার,
\(Q\) বিন্দুটি \((1)\) এর উপর অবস্থিত।
\(\therefore ax_{2}+by_{2}+c=0 ......(3)\)
\((2)\) ও \((3)\) সমাধান করে \(x_{2}\) ও \(y_{2}\) এর মান নির্ণয় করি।
\((2)\times b+(3)\times a\) এর সাহায্যে,
\(b^{2}x_{2}-b^{2}x_{1}+aby_{1}-aby_{2}+a^{2}x_{2}+aby_{2}+ac=0\)
\(\Rightarrow b^{2}x_{2}-b^{2}x_{1}+aby_{1}+a^{2}x_{2}+ac=0\)
\(\Rightarrow x_{2}(b^{2}+a^{2})=b^{2}x_{1}-aby_{1}-ac\)
\(\therefore x_{2}=\frac{b^{2}x_{1}-aby_{1}-ac}{b^{2}+a^{2}}\)
\((2)\times a-(3)\times b\) এর সাহায্যে,
\(ab^{2}x_{2}-abx_{1}+a^{2}y_{1}-a^{2}y_{2}-abx_{2}-b^{2}y_{2}-bc=0\)
\(\Rightarrow -abx_{1}+a^{2}y_{1}-a^{2}y_{2}-b^{2}y_{2}-bc=0\)
\(\Rightarrow -y_{2}(a^{2}+b^{2})=abx_{1}-a^{2}y_{1}+bc\)
\(\therefore y_{2}=-\frac{abx_{1}-a^{2}y_{1}+bc}{a^{2}+b^{2}}\)
এখন,
লম্ব দূরত্ব
\(d=PQ=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\)
\(=\sqrt{(x_{1}-\frac{b^{2}x_{1}-aby_{1}-ac}{b^{2}+a^{2}})^{2}+(y_{1}+\frac{abx_{1}-a^{2}y_{1}+bc}{a^{2}+b^{2}})^{2}}\)
\(=\sqrt{(\frac{b^{2}x_{1}+a^{2}x_{1}-b^{2}x_{1}+aby_{1}+ac}{a^{2}+b^{2}})^{2}+(\frac{a^{2}y_{1}+b^{2}y_{1}+abx_{1}-a^{2}y_{1}+bc}{a^{2}+b^{2}})^{2}}\)
\(=\sqrt{\frac{(a^{2}x_{1}+aby_{1}+ac)^{2}}{(a^{2}+b^{2})^{2}}+\frac{(b^{2}y_{1}+abx_{1}+bc)^{2}}{(a^{2}+b^{2})^{2}}}\)
\(=\sqrt{\frac{a^{2}(ax_{1}+by_{1}+c)^{2}}{(a^{2}+b^{2})^{2}}+\frac{b^{2}(ax_{1}+by_{1}+c)^{2}}{(a^{2}+b^{2})^{2}}}\)
\(=\sqrt{\frac{(ax_{1}+by_{1}+c)^{2}}{(a^{2}+b^{2})^{2}}\times (a^{2}+b^{2})}\)
\(=\sqrt{\frac{(ax_{1}+by_{1}+c)^{2}}{a^{2}+b^{2}}}\)
\(=\frac{\left|ax_{1}+by_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)
\(\therefore \) লম্ব দূরত্ব \(d=\frac{\left|ax_{1}+by_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\) ধরি,
\(P(x_{1}, y_{1})\)
\(ax+by+c=0 .........(1)\)
\(P\) হতে \((1)\) এর লম্ব দূরত্ব \(d\).
\((1)\) হতে,
\(ax+by=-c\)
\(\Rightarrow \frac{ax}{-c}+\frac{by}{-c}=\frac{-c}{-c}\) ➜Note উভয় পার্শে \(-c\) ভাগ করে
\(\Rightarrow \frac{x}{\frac{-c}{a}}+\frac{y}{\frac{-c}{b}}=1\)
\(\therefore (1)\) নং সরলরেখা অক্ষদ্বয়কে যথাক্রমে \(A(-\frac{c}{a}, 0)\) ও \(B(0, -\frac{c}{b})\) বিন্দুতে ছেদ করে।
এখন,
\(\triangle ABP=\frac{1}{2}\left|\begin{array}{c}-\frac{c}{a} \ \ \ \ 0 \ \ \ \ \ x_{1} \ \ \ -\frac{c}{a}\\ 0 \ -\frac{c}{b} \ \ \ \ \ y_{1} \ \ \ \ \ \ \ \ 0\end{array}\right|\)
\(=\frac{1}{2}\{(\frac{c^{2}}{ab}-0)+(0+\frac{cx_{1}}{b})+(0+\frac{cy_{1}}{a})\}\)
\(=\frac{1}{2}\{\frac{c^{2}}{ab}+\frac{cx_{1}}{b}+\frac{cy_{1}}{a}\}\)
\(=\frac{1}{2}\times \frac{c}{ab}\{c+ax_{1}+by_{1}\}\)
\(=\frac{c}{2ab}(ax_{1}+by_{1}+c)\)
\(AB=\sqrt{(-\frac{c}{a}-0)^{2}+(0+\frac{c}{b})^{2}}\)
\(=\sqrt{\frac{c^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}}\)
\(=\sqrt{\frac{c^{2}(a^{2}+b^{2})}{a^{2}b^{2}}}\)
\(=\frac{c}{ab}\sqrt{a^{2}+b^{2}}\)
কিন্তু,
\(\frac{1}{2}\times AB\times d=\triangle ABP\)
\(\Rightarrow \frac{1}{2}\times \frac{c}{ab}\sqrt{a^{2}+b^{2}}\times d=\frac{c}{2ab}(ax_{1}+by_{1}+c)\)
\(\Rightarrow d=\frac{c}{2ab}(ax_{1}+by_{1}+c)\times 2\times \frac{ab}{c\sqrt{a^{2}+b^{2}}}\)
\(\Rightarrow d=\frac{\left| ax_{1}+by_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

প্রমাণঃ
ধরি,
মূলবিন্দু তথা \(O(0, 0)\)
\(ax+by+c=0 ...........(1)\)
\(O(0, 0)\) হতে \((1)\) এর লম্ব দূরত্ব \(d=\frac{\left|a.0+b.0+c\right|}{\sqrt{a^{2}+b^{2}}}\)
\(\Rightarrow d=\frac{\left|0+0+c\right|}{\sqrt{a^{2}+b^{2}}}\)
\(\therefore d=\frac{\left|c\right|}{\sqrt{a^{2}+b^{2}}}\)

প্রমাণঃ
ধরি,
\(ax+by+c_{1}=0 ......(1)\)
\(ax+by+c_{2}=0 ......(2)\)
\((1)\) ও \((2)\) সরলরেখাদ্বয়ের মধ্যবর্তী লম্ব দূরত্ব \(d\)
\(p(x_{1}, y_{1})\) বিন্দুটি \((2)\) এর উপর অবস্থিত,
\(\therefore ax_{1}+by_{1}+c_{2}=0\)
\(\Rightarrow ax_{1}+by_{1}=-c_{2}...........(3)\)
আবার,
\(p(x_{1}, y_{1})\) বিন্দু হতে \((1)\) এর লম্ব দূরত্ব,
\(d=\frac{\left|ax_{1}+bx_{1}+c_{1}\right|}{\sqrt{a^{2}+b^{2}}}\)
\(=\frac{\left|-c_{2}+c_{1}\right|}{\sqrt{a^{2}+b^{2}}}\) ➜Note \((3)\) এর সাহায্যে
\(=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)
\(\therefore d=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)

প্রমাণঃ
ধরি,
\(a_{1}x+b_{1}y+c_{1}=0 ........(1)\)
\(a_{2}x+b_{2}y+c_{2}=0 ........(2)\)
\((1)\) ও \((2)\) এর অন্তর্ভুক্ত কোণের সমদ্বিখন্ডক সরলরেখাদ্বয়ের উপর যে কোনো বিন্দু \(P(x, y)\)।
\(P(x, y)\) হতে \((1)\) এর লম্ব দূরত্ব \(d_{1}=\frac{\left| a_{1}x+b_{1}y+c_{1} \right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\)
\(P(x, y)\) হতে \((2)\) এর লম্ব দূরত্ব \(d_{2}=\frac{\left| a_{2}x+b_{2}y+c_{2} \right|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)
শর্তমতে,
\(d_{1}=d_{2}\)
\(\Rightarrow \frac{\left| a_{1}x+b_{1}y+c_{1} \right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\frac{\left| a_{2}x+b_{2}y+c_{2} \right|}{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)
\(\Rightarrow \frac{ a_{1}x+b_{1}y+c_{1} }{\sqrt{a_{1}^{2}+b_{1}^{2}}}=\pm \frac{ a_{2}x+b_{2}y+c_{2} }{\sqrt{a_{2}^{2}+b_{2}^{2}}}\)
ইহাই নির্ণেয় সমদ্বিখন্ডক সরলরেখাদ্বয়ের সমীকরণ।

প্রমাণঃ
ধরি,
\(\triangle ABC\) এর \(\angle A\), \(\angle B\) কোণ দুইটির সমদ্বিখন্ডক \(AD\) ও \(BE\) পরস্পর \(I\) বিন্দুতে ছেদ করে যা ত্রিভুজের অন্তঃকেন্দ্র।
শর্তমতে,
\(\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}\)
\(\Rightarrow \frac{BD}{DC}=\frac{c}{b}\)
\(\Rightarrow BD:DC=c:b\)
\(\therefore D(\frac{bx_{2}+cx_{3}}{b+c}, \frac{by_{2}+cy_{3}}{b+c})\)
আবার,
\(\frac{AI}{ID}=\frac{AB}{BD}=\frac{AC}{CD}=\frac{AB+AC}{BD+CD}\)
\(\Rightarrow \frac{AI}{ID}=\frac{c+b}{BC}\)
\(\Rightarrow \frac{AI}{ID}=\frac{c+b}{a}\)
\(\Rightarrow AI:ID=(c+b):a\)
\(\therefore I(\frac{(b+c).\frac{bx_{2}+cx_{3}}{b+c}+a.x_{1}}{a+b+c}, \frac{(b+c).\frac{by_{2}+cy_{3}}{b+c}+a.y_{1}}{a+b+c})\)
\(\Rightarrow I(\frac{bx_{2}+cx_{3}+ax_{1}}{a+b+c}, \frac{by_{2}+cy_{3}+ay_{1}}{a+b+c})\)
\(\Rightarrow I(\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c}, \frac{ay_{1}+by_{2}+cy_{3}}{a+b+c})\).
\(\therefore \triangle ABC \) এর অন্তঃকেন্দ্র \(I(\frac{ax_{1}+bx_{2}+cx_{3}}{a+b+c}, \frac{ay_{1}+by_{2}+cy_{3}}{a+b+c})\). আবার,
\(\triangle{ABC}=\frac{1}{2}|\delta_{ABC}|=\frac{1}{2}\left|\begin{array}{c}x_{1}&x_{2}&x_{3}&x_{1}\\ y_{1}&y_{2}&y_{3}&y_{1}\end{array}\right|\)
যেখানে, \(\delta_{ABC}=\left|\begin{array}{c}x_{1}&x_{2}&x_{3}&x_{1}\\ y_{1}&y_{2}&y_{3}&y_{1}\end{array}\right|\)
এখন, \(\triangle{BIC}+\triangle{AIC}+\triangle{AIB}=\triangle{ABC}\)
\(\Rightarrow \frac{1}{2}BC\times{r}+\frac{1}{2}AC\times{r}+\frac{1}{2}AB\times{r}=\frac{1}{2}\delta_{ABC}\)
\(\Rightarrow \frac{1}{2}r(BC+AC+AB)=\frac{1}{2}|\delta_{ABC}|\)
\(\Rightarrow r(BC+AC+AB)=|\delta_{ABC}|\)
\(\Rightarrow r=\frac{|\delta_{ABC}|}{BC+AC+AB}\)
\(\therefore r=\frac{|\delta_{ABC}|}{a+b+c}\)

প্রমাণঃ
ধরি,
\(ax+by+c=0 ............(1)\)
\((1)\) এর সমান্তরাল যে কোনো সরলরেখার সমীকরণ,
\(ax+by+k=0 ............(2)\) ➜Note \(k\) যে কোন বাস্তব সংখ্যা।
\((1)\) ও \((2)\) এর মধ্যবর্তী দূরত্ব \(=\frac{|k-c|}{\sqrt{a^{2}+b^{2}}}\)
শর্তমতে,
\(\frac{|k-c|}{\sqrt{a^{2}+b^{2}}}=d\)
\(\Rightarrow \pm \frac{k-c}{\sqrt{a^{2}+b^{2}}}=d\)
\(\Rightarrow \frac{k-c}{\sqrt{a^{2}+b^{2}}}=\pm d\)
\(\Rightarrow k-c=\pm d \sqrt{a^{2}+b^{2}}\)
\(\therefore k=c\pm d \sqrt{a^{2}+b^{2}}\)
\(k\) এর মান \((2)\) এ বসিয়ে,
\(ax+by+c\pm d \sqrt{a^{2}+b^{2}}=0\)
ইহাই নির্ণেয় সরলরেখার সমীকরণ।

প্রমাণঃ
ধরি,
\(ax+by+c=0 ............(1)\)
\((1)\) এর উপর লম্ব যে কোনো সরলরেখার সমীকরণ,
\(bx-ay+k=0 ............(2)\) ➜Note \(k\) যে কোন বাস্তব সংখ্যা।
\((x_{1}, y_{1})\) বিন্দু হতে \((2)\) এর লম্ব দূরত্ব \(=\frac{|bx_{1}-ay_{1}+k|}{\sqrt{a^{2}+b^{2}}}\)
শর্তমতে,
\(\frac{|bx_{1}-ay_{1}+k|}{\sqrt{a^{2}+b^{2}}}=d\)
\(\Rightarrow |bx_{1}-ay_{1}+k|=d\sqrt{a^{2}+b^{2}}\)
\(\Rightarrow bx_{1}-ay_{1}+k=\pm d\sqrt{a^{2}+b^{2}}\)
\(\therefore k=ay_{1}-bx_{1}\pm d\sqrt{a^{2}+b^{2}}\)
\(k\) এর মান \((2)\) এ বসিয়ে,
\(bx-ay+ay_{1}-bx_{1}\pm d\sqrt{a^{2}+b^{2}}=0\)
\(\therefore b(x-x_{1})-a(y-y_{1})\pm d\sqrt{a^{2}+b^{2}}=0\)
ইহাই নির্ণেয় সরলরেখার সমীকরণ।