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.

#### Solution

We know that the equations of two lines passing through a point \[\left( x_1 , y_1 \right)\] and making an angle \[\alpha\] with the given line y = m^{'}x + c are \[y - y_1 = \frac{m^{\prime}\pm \tan\alpha}{1 \mp m^{\prime} \tan\alpha}\left( x - x_1 \right)\]

Here,

\[x_1 = h, y_1 = k, \alpha = \tan^{- 1} m, m^{\prime}= m\]

So, the equations of the required lines are

\[y - k = \frac{m + m}{1 - m^2}\left( x - h \right) and y - k = \frac{m - m}{1 + m^2}\left( x - h \right)\]

\[ \Rightarrow y - k = \frac{2m}{1 - m^2}\left( x - h \right) \text { and } y - k = 0\]

\[ \Rightarrow \left( y - k \right)\left( 1 - m^2 \right) = 2m\left( x - h \right)\text { and } y = k\]