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 $\mathrm{\alpha }$ 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,
$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}, \\ 3x+y-5=0 \\ \Rightarrow y=-3x+5 \\ \mathrm{Comparing}\mathrm{this}\mathrm{equation}\mathrm{with}\mathit{}y=mx+c \\ \mathrm{we}\mathrm{get}, \\ m=-3 \end{gathered}$$
$x_1=2,y_1=3,\mathrm{\alpha }={45}^{\circ },m=-3$.

So, the equations of the required lines are

$$\begin{gathered} y-3=\frac{{\displaystyle -3+\tan {45}^{\circ }}}{{\displaystyle 1+3\tan {45}^{\circ }}}\left(x-2\right)\mathrm{and}y-3=\frac{{\displaystyle -3-\tan {45}^{\circ }}}{{\displaystyle 1-3\tan {45}^{\circ }}}\left(x-2\right) \\ \Rightarrow y-3=\frac{{\displaystyle -3+1}}{{\displaystyle 1+3}}\left(x-2\right)\mathrm{and}y-3=\frac{{\displaystyle -3-1}}{{\displaystyle 1-3}}\left(x-2\right) \\ \Rightarrow y-3=\frac{{\displaystyle -1}}{{\displaystyle 2}}\left(x-2\right)\mathrm{and}y-3=2\left(x-2\right) \\ \Rightarrow x+2y-8=0\mathrm{and}2x-y-1=0 \end{gathered}$$