Question

Find the equations to the straight lines which pass through the point (h, k) and are inclined at angle tan⁻¹ m to the straight line y = mx + c.

Answer 1

We know that the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = m^'x + c are

$y-y_1=\frac{m^{\text{'}}\pm \tan \alpha }{1\mp m^{\text{'}}\tan \alpha }\left(x-x_1\right)$

Here,
$x_1=h,y_1=k,\mathrm{\alpha }={\tan }^{-1}m,m^{\text{'}}=m$.

So, the equations of the required lines are

$$\begin{gathered} y-k=\frac{{\displaystyle m}+m}{1-{\displaystyle m^2}}\left(x-h\right)\mathrm{and}y-k=\frac{{\displaystyle m-}m}{1+{\displaystyle m^2}}\left(x-h\right) \\ \Rightarrow y-k=\frac{2{\displaystyle m}}{1-{\displaystyle m^2}}\left(x-h\right)\mathrm{and}y-k=0 \\ \Rightarrow \left(y-k\right)\left(1-m^2\right)=2m\left(x-h\right)\mathrm{and}y=k \end{gathered}$$