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I have two lines

$y-8=\dfrac{-1}{7}(x+6)$

$y-8=\dfrac{-1}{2}(x+6)$

They both intersect at point $(-6,8)$.

I'm trying to find the slope of the line that bisects these two lines.

In this question, there was an equation for the line of the angle bisector. Equation of angle bisector, given the equations of two lines in 2D

I'm new to the math forum, so I'm not sure how to copy the equation to post in this question with the pretty formatting.

$\dfrac{(a_1 x+b_1 y-c_1)}{\sqrt{a_1^2+b_1^2}} = \dfrac{(a_2x+b_2y-c_2)}{\sqrt{a_2^2+b_2^2}} $

It sets the two equations equal to each other and divides both by basically the distance formula of the two coefficients of each equation. However, this equation does not give a simple way to know or find the slope of the bisector.

I was wondering if there was a simpler way to write the equation of the bisector line in terms of the new slope. I did find a similar question here. Most accurate linear approximation for two lines

It gives the 3rd slope as the tangent of $\dfrac{(Angle1 + Angle2)}{2}$. However, when I try to graph this, the new line does not bisect my two lines.

Thanks,

Community
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Christopher Gaston
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2 Answers2

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Are you converting the slopes of the two lines to angles first? (Use the $\arctan$ function). Then you average them and take the tangent.

$$\theta_1 = \arctan(m_1)$$ $$\theta_2 = \arctan(m_2)$$ $$\theta_b = \frac{(\theta_1 + \theta_2)}{2}$$ $$m_b = \tan(\theta_b)$$

Your final answer for $m_b$ should be about -0.31, giving you $y-8=-0.31(x+6)$.

(If this is what you did, sorry. It's working for me.)

a52
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  • The exact value is $\frac {\pm13-5\sqrt{10}}{9}$ – David Quinn Nov 09 '16 at 22:24
  • Not sure why I was getting something different, but when I redid it just now, I got the same thing. So basically in terms of the new slope the equation would be: $m_3=tan(\dfrac{tan^{-1}(m_1)+tan^{-1}(m_2)}{2})$ or $m_3=-tan(\dfrac{tan^{-1}(m_1)+tan^{-1}(m_2)}{2})^{-1}$ if I wanted the perpendicular slope. Thanks. – Christopher Gaston Nov 10 '16 at 15:53
  • David Quinn, how did you find the exact value? – Christopher Gaston Nov 10 '16 at 16:24
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bisector

Given two lines, you have two possible bisecting lines : that of the obtuse angle and that of the acute angle.
But the two angles could also be rect and thus equal.
Take the unit tangent vector of each line, or otherwise the unit normal vectors.
Taking their dot product, you can determine if they individuate the acute or the obtuse angle. In case they encompass an obtuse angle, and you want to determine the bisector of the acute one, invert one the vectors. Then sum the vectors (subtract) : the resultant will be a vector (not unitary) parallel / normal to the bisecting line. Of course, once you have the vector you also have the angle it makes with the $x$ axis, etc.

In the case you present $$ \frac{{y - 8}} {{ - 1}} = \frac{{x + 6}} {7}\quad \frac{{y - 8}} {{ - 1}} = \frac{{x + 6}} {2} $$ the two unit parallel vectors are: $$ \mathbf{t}_\mathbf{1} = \left( {\frac{7} {{\sqrt {50} }},\;\frac{{ - 1}} {{\sqrt {50} }}} \right)\quad \mathbf{t}_\mathbf{2} = \left( {\frac{2} {{\sqrt 5 }},\;\frac{{ - 1}} {{\sqrt 5 }}} \right) $$ for which we have: $$ \mathbf{t}_\mathbf{1} \cdot \mathbf{t}_\mathbf{2} = \frac{{15}} {{\sqrt {250} }} = \frac{3} {{\sqrt {10} }} > 0\quad \phi _{\,t,\,t} = \arccos \left( {\frac{3} {{\sqrt {10} }}} \right) \approx 18.435^\circ $$ and thus are relative to the acute angle.

The two (not normalized) parallel vectors corresponding to the bisecting lines are therefore: $$ \mathbf{b}_{\,\mathbf{1}\mathbf{,}\,\mathbf{2}} = \left( {7 \pm 2\sqrt {10} ,\; - 1 \mp \sqrt {10} } \right) $$

G Cab
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