Question

Find the equation of a straight line whose inclination is ${60}^0$ and which passes through the point $\left(0,-3\right)$.

Answer 1

Calculation the equation of a straight line:

The slope of a line is defined as $m=\tan \theta$.

The equation of line in point slope form is given by $y-y_1=m\left(x-x_1\right)$, where $m$ is the slope of the line and it passes through the point $(x_1,y_1)$.

The slope of the given line is $\tan {60}^0$ i.e. $\sqrt{3}$.

The points is $(0,-3)$.

Therefore, $x_1=0,y_1=-3$ and $m=\sqrt{3}$.

The equation of the line is

$$\begin{gathered} y-(-3)=\sqrt{3}\left(x-0\right) \\ \Rightarrow y+3=\sqrt{3}x \\ \therefore \sqrt{3}x-y-3=0(subtractingy+3frombothside) \end{gathered}$$

Hence, the equation straight line is $\sqrt{3}x-y-3=0$