Question

# Find the equation of the straight line passing through (3, -2) and making an angle of 60° with the positive direction of y-axis.

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Hint: Solve by finding the slope of the line. Then, use the formula for the equation of line passing through the point and having a given slope to find its equation. Solve by finding the slope of the line. Then, use the formula for the equation of line passing through the point and having a given slope to find its equation. The equation of a point that passes through a point (a, b) and has a slope m is given as $y-b = m(x-a)$.

We know that the slope of a line is the tangent of the angle that it makes with the positive direction of x-axis.
If m is the slope of the line and $\theta$ is the angle made with the positive direction of x-axis, then we have the relation as follows:
$m = \tan \theta ............(1)$
It is given that the line makes an angle 60° with the positive direction of the y-axis. We know that the angle between positive direction of y-axis and positive direction of x-axis is 90°, then the angle made by the line with the positive direction of the x-axis is 90° minus the angle made with the positive direction of y-axis.
$\theta = 90^\circ - 60^\circ$
$\theta = 30^\circ .............(2)$
Now using equation (2) in equation (1), we get:
$\Rightarrow$ $m = \tan 30^\circ$
$\Rightarrow$ $m = \dfrac{1}{{\sqrt 3 }}...........(3)$
It is given that the line passes through the point (3, -2) and we found the value of slope to be $\dfrac{1}{{\sqrt 3 }}$.
The equation of a point that passes through a point (a, b) and has a slope m is given as follows:
$y - b = m(x - a)$
Substituting the values in the above equation, we have:
$\Rightarrow$ $y - ( - 2) = \dfrac{1}{{\sqrt 3 }}(x - 3)$
$\Rightarrow$ $y + 2 = \dfrac{1}{{\sqrt 3 }}(x - 3)$
Taking $\sqrt 3$ to the other side, we have:
$\Rightarrow$ $\sqrt 3 (y + 2) = x - 3$
Simplifying the equation, we obtain:
$\Rightarrow$ $x - \sqrt 3 y - 2\sqrt 3 - 3 = 0$
$\Rightarrow$ $x = \sqrt 3 y + 2\sqrt 3 + 3$
Hence, the required equation is $x = \sqrt 3 y + 2\sqrt 3 + 3$.

Note: Do not use the given angle to calculate the tangent to find the slope directly, first, you need to find the angle made by the line with the positive direction of x-axis, then proceed with the solution.