Question

Find the equations to the straight lines passing through the point (2, 3) and inclined at an angle of ${45}^{\circ }$ to the line $3x+y-5=0$

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

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

Here,

Equation of the given line is,

$3x+y-5=0$

$\Rightarrow y=-3x+5$

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

$y-3=\frac{-3+tan{45}^{\circ }}{1+3tan{45}^{\circ }}(x-2)$ and $y-3$

$=\frac{-3-tan{45}^{\circ }}{1-3tan{45}^{\circ }}(x-2)$

$\Rightarrow y-3=\frac{-3+1}{1+3}(x-2)$ and $y-3=\frac{-3-1}{1-3}(x-2)$

$\Rightarrow y-3=\frac{-1}{2}(x-2)$ and $y-3=2(x-2)$

$\Rightarrow x+2y-8=0$ and $2x-y-1=0$

$y-3=\frac{-3+tan{45}^{\circ }}{1+3tan{45}^{\circ }}(x-2)$ and $y-3=\frac{-3-tan{45}^{\circ }}{1-3tan{45}^{\circ }}(x-2)$