Question

Find the equations of the straight lines passing through (2, −1) and making an angle of 45° with the line 6x + 5y − 8 = 0.

Answer 1

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 = mx + c are

\[y - y_1 = \frac{m \pm \tan\alpha}{1 \mp m\tan\alpha}\left( x - x_1 \right)\]

Here,

Equation of the given line is,

\[6x + 5y - 8 = 0\]

\[ \Rightarrow 5y = - 6x + 8\]

\[ \Rightarrow y = - \frac{6}{5}x + \frac{8}{5}\]

\[\text { Comparing this equation with } y = mx + c\]

we get, 

\[m = - \frac{6}{5}\] 

\[x_1 = 2, y_1 = - 1, \alpha = {45}^\circ , m = - \frac{6}{5}\]

So, the equations of the required lines are

\[y + 1 = \frac{- \frac{6}{5} + \tan {45}^\circ}{1 + \frac{6}{5}\tan {45}^\circ}\left( x - 2 \right) \text { and }y + 1 = \frac{- \frac{6}{5} - \tan {45}^\circ}{1 - \frac{6}{5}\tan {45}^\circ}\left( x - 2 \right)\]

\[ \Rightarrow y + 1 = \frac{- \frac{6}{5} + 1}{1 + \frac{6}{5}}\left( x - 2 \right) \text { and } y + 1 = \frac{- \frac{6}{5} - 1}{1 - \frac{6}{5}}\left( x - 2 \right)\]

\[ \Rightarrow y + 1 = \frac{- 1}{11}\left( x - 2 \right) \text { and } y + 1 = \frac{- 11}{- 1}\left( x - 2 \right)\]

\[ \Rightarrow x + 11y + 9 = 0\text { and } 11x - y - 23 = 0\]