# Find the equation of a line whose inclination with the x - axis is

Question:

Find the equation of a line whose inclination with the $x$-axis is $150^{\circ}$ and which passes through the point $(3,-5)$.

Solution:

As angle is given so we have to find slope first give by m = tanθ

$m=\tan 150^{\circ}$

$\mathrm{m}=\tan \left(180^{\circ}-30^{\circ}\right) \Rightarrow-\tan 30^{\circ}=-\frac{1}{\sqrt{3}}\left(\tan \left(180^{\circ}-\theta\right)\right.$ is in II quadrant, $\tan x$ is negative)

Now the line is passing through the point $(3,-5)$. Using the slope - intercept form of the equation of the line, we will find the intercept

$y=m x+c$..................(1)

$-5=-\frac{1}{\sqrt{3}}(3)+\mathrm{c} \Rightarrow \mathrm{c}=-5+\sqrt{3}$

Putting the value of c in equation (1),we have

$y=-\frac{1}{\sqrt{3}} x+(-5+\sqrt{3})$

$x+(\sqrt{3}) y+5 \sqrt{3}-3=0$

So, required equation of line is $\mathrm{x}+(\sqrt{3}) \mathrm{y}+5 \sqrt{3}-3=0$.

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