Question

Find the equations of the straight lines passing through (2, -1) and making an angle of ${45}^{\circ }$ with the line $6x+5y-8=0$.

Answer 1

We know that the equations of two lines passing through a point $(x_1,y_1)$ and making an angle $\alpha$ with the given line $y=mx+c$ are

$y-y_1=\frac{m\pm tan\alpha }{1\mp mtan\alpha }(x-x_1)$

Here,

Equation of the given line is,

$6x+5y-8=0$

$\Rightarrow 5y=-6x+8$

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

Comparing this eqution with $y=mx+c$ we get,

$m=-\frac{6}{5}$

$x_1=2$, $y=-1$, $\alpha ={45}^{\circ }$, $m=-\frac{6}{5}$

So, the equations of the required lines are

$y+1=\frac{-\frac{6}{5}+tan{45}^{\circ }}{1+\frac{6}{5}tan{45}^{\circ }}(x-2)$ and $y+1=\frac{-\frac{6}{5}-tan{45}^{\circ }}{1-\frac{6}{5}tan{45}^{\circ }}(x-2)$

$\Rightarrow y+1=\frac{-\frac{6}{5}+1}{1+\frac{6}{5}}(x-2)$

$\Rightarrow x+11y+9=0$ and $11x-y-23=0$