babyagelesszj

2022-07-07

Two lines: ${a}_{1}x+{b}_{1}y+{c}_{1}=0$ and ${a}_{2}x+{b}_{2}y+{c}_{2}=0$ are given. I know that the equation of its bisectors is $\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{\left({a}_{1}^{2}+{b}_{1}^{2}\right)}}=±\frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{\left({a}_{2}^{2}+{b}_{2}^{2}\right)}}$ But I intend to find which one is the obtuse angle bisector and which one is the acute angle bisector. I want to find a general formula Assuming ${c}_{1},{c}_{2}$ both are of same sign, I know if ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0$ and if we take the positive sign we get the obtuse angle bisector and vice versa. But I want to prove it using general equation of line, I tried to find the angle between bisector and original line i.e. $tan\theta =\frac{{m}_{1}-{m}_{2}}{1+{m}_{1}{m}_{2}}$ and then if it is greater than one it will be of obtuse angle but calculations are tough if we use general equation of line. May anyone give a simple proof of the following statement: "Assuming ${c}_{1},{c}_{2}$ both are of same sign IF ${a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0$ then if we take positive sign we get the obtuse angle bisector".

Dobermann82

Expert

We have two lines :
${L}_{1}:{a}_{1}x+{b}_{1}y+{c}_{1}=0,\phantom{\rule{1em}{0ex}}{L}_{2}:{a}_{2}x+{b}_{2}y+{c}_{2}=0$
and the angle bisectors :
${L}_{±}:\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}=±\frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}$
If we let $\theta$ be the (smaller) angle between ${L}_{+}$ and ${L}_{1}$, then we have
$\mathrm{cos}\theta =\frac{|{a}_{1}\left(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}\right)+{b}_{1}\left(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}\right)|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{{\left(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}\right)}^{2}+{\left(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}\right)}^{2}}}$
$=\frac{|\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{2-2\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{\left({a}_{1}^{2}+{b}_{1}^{2}\right)\left({a}_{2}^{2}+{b}_{2}^{2}\right)}}}}×\frac{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}=\sqrt{\frac{1-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{\left({a}_{1}^{2}+{b}_{1}^{2}\right)\left({a}_{2}^{2}+{b}_{2}^{2}\right)}}}{2}}$
Hence, we can see that

as desired.

(Note that "${c}_{1},{c}_{2}$ both are of same sign" is irrelevant.)

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