babyagelesszj

Answered

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 $\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})}}=\pm \frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{({a}_{2}^{2}+{b}_{2}^{2})}}$ 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".

Answer & Explanation

Dobermann82

Expert

2022-07-08Added 15 answers

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 :

${L}_{\pm}:\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}=\pm \frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}$

If we let $\theta $ be the (smaller) angle between ${L}_{+}$ and ${L}_{1}$, then we have

$\mathrm{cos}\theta =\frac{|{a}_{1}(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})+{b}_{1}(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{{(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})}^{2}+{(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})}^{2}}}$

$=\frac{|\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{2-2\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})({a}_{2}^{2}+{b}_{2}^{2})}}}}\times \frac{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}=\sqrt{\frac{1-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})({a}_{2}^{2}+{b}_{2}^{2})}}}{2}}$

Hence, we can see that

$\begin{array}{rl}{a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta <1/\sqrt{2}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\theta >{45}^{\circ}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{L}_{+}\text{is the obtuse angle bisector}\end{array}$

as desired.

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

${L}_{1}:{a}_{1}x+{b}_{1}y+{c}_{1}=0,\phantom{\rule{1em}{0ex}}{L}_{2}:{a}_{2}x+{b}_{2}y+{c}_{2}=0$

and the angle bisectors :

${L}_{\pm}:\frac{{a}_{1}x+{b}_{1}y+{c}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}=\pm \frac{{a}_{2}x+{b}_{2}y+{c}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}$

If we let $\theta $ be the (smaller) angle between ${L}_{+}$ and ${L}_{1}$, then we have

$\mathrm{cos}\theta =\frac{|{a}_{1}(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})+{b}_{1}(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{{(\frac{{a}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{a}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})}^{2}+{(\frac{{b}_{1}}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}-\frac{{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}})}^{2}}}$

$=\frac{|\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{{a}_{2}^{2}+{b}_{2}^{2}}}|}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}\sqrt{2-2\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})({a}_{2}^{2}+{b}_{2}^{2})}}}}\times \frac{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}{2\frac{1}{\sqrt{{a}_{1}^{2}+{b}_{1}^{2}}}}=\sqrt{\frac{1-\frac{{a}_{1}{a}_{2}+{b}_{1}{b}_{2}}{\sqrt{({a}_{1}^{2}+{b}_{1}^{2})({a}_{2}^{2}+{b}_{2}^{2})}}}{2}}$

Hence, we can see that

$\begin{array}{rl}{a}_{1}{a}_{2}+{b}_{1}{b}_{2}>0& \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta <1/\sqrt{2}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}\theta >{45}^{\circ}\\ & \phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{L}_{+}\text{is the obtuse angle bisector}\end{array}$

as desired.

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

Most Popular Questions