Question

Find the equations to the straight lines passing through the point (2, 3) and inclined at and angle of 45° to the line 3x + y − 5 = 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,

\[3x + y - 5 = 0\]

\[ \Rightarrow y = - 3x + 5\]

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

we get,

\[m = - 3\]

\[x_1 = 2, y_1 = 3, \alpha = {45}^\circ , m = - 3\]

So, the equations of the required lines are

\[y - 3 = \frac{- 3 + \tan {45}^\circ}{1 + 3\tan {45}^\circ}\left( x - 2 \right) \text { and } y - 3 = \frac{- 3 - \tan {45}^\circ}{1 - 3\tan {45}^\circ}\left( x - 2 \right)\]

\[ \Rightarrow y - 3 = \frac{- 3 + 1}{1 + 3}\left( x - 2 \right) \text { and } y - 3 = \frac{- 3 - 1}{1 - 3}\left( x - 2 \right)\]

\[ \Rightarrow y - 3 = \frac{- 1}{2}\left( x - 2 \right) \text { and } y - 3 = 2\left( x - 2 \right)\]

\[ \Rightarrow x + 2y - 8 = 0 \text { and } 2x - y - 1 = 0\]