Lines $L_1$ ($a_1x+b_1y+c_1$) and $L_2$ ($a_2x+b_2y+c_2$) intersect at a point $P$ and subtend an angle $\theta$ at $P$.

Another line $L$ makes the same angle $\theta$ with $L_1$ at $P$. Find the equation of L.

$$$$All I could think of was to use the concepts of Family of Lines and Angle Bisectors (I couldn't think of how though). We already have the equations of two lines passing through a fixed point $P$. Thus, the equation of $L$ must be of the form $$(a_1x+b_1y+c_1)+\lambda(a_2x+b_2y+c_2)=0$$

However, at this point, I got stuck since I was unable to calculate the value of $\lambda$.

$$$$Any help with this question would be greatly appreciated. Many thanks in anticipation!


If $L_1,L_2$ have inclinations $ \alpha , \beta$ to x-axis, then the required intermediate angle has inclination. $$ \alpha + \lambda \beta $$

It is just a hunch, please check.


Trying to show how the various slopes play out.

The set of lines $ L_1,L_2$ have a common point of concurrency. I.e., they all pass through a common point of intersection.

$$ L_1: \, y= m_1x + c_1;\quad L_2: \, y= m_2x + c_2 $$

A concurrent set forms where $ \lambda $ can be varied through the join of lines $ L_1=0, L_2=0$ as:

$$ \lambda L_1+ (1-\lambda) L_2=0 $$

As examples for concurrent lines we take $ \lambda= \pm \dfrac12 $ and picture two resulting lines in the graph:

Chosen lines $$ y= mx+c;\, m_1=-1, m_2= \sqrt3, c_1=1, c_2=2\,$$

Concurrent Lines

Common point of intersection $P$

$$ (x_I,y_I)= -\frac{\sqrt3-1}{2}, +\frac{\sqrt3+1}{2} $$

After this is understood we need only consider left hand side part going to zero for obtaining the internal and external bisectors:

$$ \dfrac{L_1}{\sqrt{1+m_1^2}} =\dfrac{L_2}{\sqrt{1+m_2^2}} ; $$

Hope now you can address the main question. (Finding $\lambda$ when extra angle/slope is added)


So this new line, L, crosses line L1 on the other side of L1. If line L1 makes angle $\phi$ with the x-axis then line L2 makes angle $\phi+ \theta$ or $\phi- \theta$. In the first case, L must make angle $\phi- \theta$ with the x-axis and in the second, angle $\theta+ \phi$.

  • $\begingroup$ In your first case, the angle of inclination of L (measured from the positive x-axis in the anti-clockwise direction) is $\pi – (\theta - \phi)$. Of course, $m_L = \tan [\pi – (\theta - \phi)] = … = \tan (\phi – \theta)$ $\endgroup$ – Mick Oct 27 '15 at 4:33

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