The equation of the straight line cutting off | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The equation of a line in slope-intercept form is given by $y = mx + c$.Given that the line is inclined at $30^\circ$ to the positive direction of

Vedclass · Accepted Answer

$\sqrt{3}y - x + 2\sqrt{3} = 0$

Vedclass · Answer

(D) The equation of a line in slope-intercept form is given by $y = mx + c$.Given that the line is inclined at $30^\circ$ to the positive direction of the $x$-axis,the slope $m = 	an(30^\circ) = \frac{1}{\sqrt{3}}$.The line cuts off an intercept of $2$ from the negative direction of the $y$-axis,so $c = -2$.Substituting these values into the equation,we get $y = \frac{1}{\sqrt{3}}x - 2$.Multiplying both sides by $\sqrt{3}$,we get $\sqrt{3}y = x - 2\sqrt{3}$.Rearranging the terms,we obtain $\sqrt{3}y - x + 2\sqrt{3} = 0$.

The equation of the straight line cutting off an intercept of $2$ from the negative direction of the $y$-axis and inclined at $30^\circ$ to the positive direction of the $x$-axis is:

Similar Questions

The ends of the base of an isosceles triangle are at $(2a, 0)$ and $(0, a)$. The equation of one side is $x = 2a$. The equation of the other side is

The equation of a line passing through $(3, -4)$ and perpendicular to the line $3x + 4y = 5$ is

If the slope of a line is $2$ and it makes an intercept of $-4$ on the $y$-axis,then its equation is:

If a straight line is at a distance of $10$ units from the origin and the perpendicular drawn from the origin to it makes an angle of $\frac{\pi}{4}$ with the negative $X$-axis in the negative direction,then the equation of that line is:

If the coordinates of the points $A, B, C, D$ are $(a, b), (a', b'), (-a, b)$ and $(a', -b')$ respectively,then the equation of the line bisecting the line segments $AB$ and $CD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module