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Question Number 78820 by mathocean1 last updated on 20/Jan/20

$$\mathrm{please}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{fomula}\mathrm{to}$$$$\mathrm{determinate}\mathrm{the}\mathrm{equations}\mathrm{of}$$$$\mathrm{bissectors}\mathrm{in}\mathrm{triangle}?$$

Answered by MJS last updated on 21/Jan/20

$$\mathrm{bisector}\mathrm{between}2\mathrm{lines}$$$$l_1:y=a_{1}x+c_1$$$$l_2:y=a_{2}x+c_2$$$$\mathrm{intersection}:$$$$I=\left(\begin{array}{c}\frac{c_2-c_1}{a_1-a_2}\\ \frac{a_1{c}_2-a_2{c}_1}{a_1-a_2}\end{array}\right)$$$$\mathrm{direction}\mathrm{vectors}:$$$$v_1=\left(\begin{array}{c}1\\ a_1\end{array}\right);v_2=\left(\begin{array}{c}1\\ a_2\end{array}\right)$$$$\mathrm{normalized}:$$$$v_1=\left(\begin{array}{c}\frac{1}{\sqrt{1+a_1^2}}\\ \frac{a_1}{\sqrt{1+a_1^2}}\end{array}\right);v_2=\left(\begin{array}{c}\frac{1}{\sqrt{1+a_2^2}}\\ \frac{a_1}{\sqrt{1+a_2^2}}\end{array}\right)$$$$\mathrm{direction}\mathrm{vectors}\mathrm{of}\mathrm{bisectors}$$$$v_{b1}=v_1+v_2=\left(\begin{array}{c}p\\ q\end{array}\right)$$$$v_{b2}=\left(\begin{array}{c}q\\ -p\end{array}\right)$$$$\mathrm{bisectors}$$$$l_{b1}:X=I+{tv}_{b1}$$$$l_{b2}:X=I+{tv}_{b2}$$

Commented by jagoll last updated on 21/Jan/20

$$\mathrm{good}\mathrm{sir}$$

Commented by mathocean1 last updated on 21/Jan/20

$$\mathrm{hello}$$$$\mathrm{is}\mathrm{it}\mathrm{the}\mathrm{same}\mathrm{thing}\mathrm{of}$$$$\mathrm{bissectrice}\mathrm{of}\mathrm{angles}?$$

Commented by MJS last updated on 21/Jan/20

$$\mathrm{of}\mathrm{course}.\mathrm{what}\mathrm{is}\mathrm{an}\mathrm{angle}\mathrm{without}\mathrm{lines}?$$

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