Hello GTK X Hunter Quote:

Originally Posted by

**GTK X Hunter** Let $\displaystyle l$ be a line through $\displaystyle (x_0,y_0)$ cutting the line $\displaystyle k: y=mx+c$ and forming an angle $\displaystyle \alpha$. Prove that the equation of the line $\displaystyle l$ is

$\displaystyle y-y_0=\frac{m-\tan\alpha}{1+m\tan\alpha}(x-x_0)$ or

$\displaystyle y-y_0=\frac{\tan\alpha-m}{1+m\tan\alpha}(x-x_0)$

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 $\displaystyle m_1$ passing through the point $\displaystyle (x_0,y_0)$ is$\displaystyle y-y_0=m_1(x-x_0)$

This makes an angle $\displaystyle \theta$, say, with the direction of the $\displaystyle x$-axis, where $\displaystyle \tan\theta = m_1$.

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

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

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

$\displaystyle =\frac{m\pm\tan\alpha}{1\mp m\tan\alpha}$

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

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

Grandad