Question

The equation of the line making an angle ${60^0}$ with x-axis and y-intercept $\dfrac{{ - 2}}{5}$ is:

Answer 1

Hint : The general equation of a straight line is $y = mx + c$ , where $m$ is the slope or the gradient, $c$ is the y-intercept. In this question the angle which the line is making with the x-axis and the y-intercept is given so we will first find the slope of the line using the given angle and then we will substitute these values in the general equation of a straight line to find the equation of the straight line.

Complete step-by-step answer :
The y-intercept of the line is given as, $c = - \dfrac{2}{5}$
The line is making an angle of $\theta = {60^ \circ }$ with the x-axis so the slope of the line which is given by the formula $m = \tan \theta$ will be
$m = \tan {60^ \circ } = \sqrt 3 {\text{ }}\left( {\because \tan {{60}^ \circ } = \sqrt 3 } \right)$
So, the slope of the line is given as $m = \sqrt 3$
Now, we know the general equation of a straight line is $y = mx + c$ , so we will substitute the values of the slope and the y-intercept in the equation, hence the equation of the straight line will be

$$y = \left( {\sqrt 3 } \right)x - \dfrac{2}{5} \\\ \Rightarrow y = \sqrt 3 x - \dfrac{2}{5} \;$$

Therefore, the required equation of the line whose slope is $m = \sqrt 3$ and the y-intercept $c = - \dfrac{2}{5}$ is
$y = \sqrt 3 x - \dfrac{2}{5}$
So, the correct answer is “ $y = \sqrt 3 x - \dfrac{2}{5}$ ”.

Note : Here, the y-intercept of the line is $c = - \dfrac{2}{5}$ , the y-intercept means that the straight line cuts the y-axis at the point $- \dfrac{2}{5}$ which is in the negative side of the y-axis on an x-y graph plane and the line will move in the positive x-axis direction since the slope of the line is positive . We can plot the graph as