Find the equation of a straight line whose inclination is $60 °$ and $y - i n t e r c e p t$ is $- 3$ .

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Solution

Step 1: Finding the slope of the line using the given inclination.
Given inclination is $60 °$ .
$∴$ Slope of the line is given by,
$m = \tan (θ)$
$\Rightarrow m = \tan (60 °)$
$\Rightarrow m = \sqrt{3}$
Step 2: Finding the equation of the line using slope-intercept form.
The general equation of a line having slope $m$ and $y - i n t e r c e p t$ as $c$ is given by
$y = m x + c$
The equation of the line having slope $m = \sqrt{3}$ and $y - i n t e r c e p t$ , $c = - 3$ will be
$\Rightarrow y = \sqrt{3} x + (- 3)$
$\Rightarrow \sqrt{3} x - y - 3 = 0$
Hence, the required equation is $\sqrt{3} x - y - 3 = 0$

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