# crossing over line

• Oct 29th 2009, 09:38 AM
GTK X Hunter
crossing over line
Let $l$ be a line through $(x_0,y_0)$ cutting the line $k: y=mx+c$ and forming an angle $\alpha$. Prove that the equation of the line $l$ is
$y-y_0=\frac{m-\tan\alpha}{1+m\tan\alpha}(x-x_0)$ or
$y-y_0=\frac{\tan\alpha-m}{1+m\tan\alpha}(x-x_0)$
• Oct 30th 2009, 06:06 AM
Hello GTK X Hunter
Quote:

Originally Posted by GTK X Hunter
Let $l$ be a line through $(x_0,y_0)$ cutting the line $k: y=mx+c$ and forming an angle $\alpha$. Prove that the equation of the line $l$ is
$y-y_0=\frac{m-\tan\alpha}{1+m\tan\alpha}(x-x_0)$ or
$y-y_0=\frac{\tan\alpha-m}{1+m\tan\alpha}(x-x_0)$

I think you have an error in the second of these two equations. Here is what I believe to be the correct result.

The equation of the line with gradient $m_1$ passing through the point $(x_0,y_0)$ is
$y-y_0=m_1(x-x_0)$
This makes an angle $\theta$, say, with the direction of the $x$-axis, where $\tan\theta = m_1$.

The line $y = mx + c$ makes an angle $\phi$, say, with the $x$-axis, where $\tan\phi = m$.

Then, depending on which side of the line $y = mx + c$ the angle $\alpha$ is made:
$\theta= \phi\pm\alpha$ (See the attached diagram. The lines shown are not necessarily the actual lines, but simply parallel to them through a single point.)

$\Rightarrow m_1=\tan\theta = \tan(\phi\pm\alpha)= \frac{\tan\phi \pm\tan\alpha}{1\mp\tan\phi\tan\alpha}$

$=\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}$
So the two gradients are $\frac{m+\tan\alpha}{1-\tan\alpha}$ and $\frac{m-\tan\alpha}{1+\tan\alpha}$.

The error in the question can be easily verified, because the gradient of one of the the given lines is $(-1) \times$ the gradient of the other. Therefore they make equal angles above and below the $x$-axis. They should, of course, make equal angles above and below the line $y = mx + c$.