Suppose a line makes an angle of 120^{circ} w | vedclass

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

Question

(B) The line makes an angle of $120^{\circ}$ with the positive direction of the $X$-axis. The slope of the line is $m = 	an(120^{\circ}) = -\sqrt{3}$

Vedclass · Accepted Answer

$\sqrt{3}x + y = 8$

Vedclass · Answer

(B) The line makes an angle of $120^{\circ}$ with the positive direction of the $X$-axis. The slope of the line is $m = 	an(120^{\circ}) = -\sqrt{3}$.Let the normal to the line make an angle $\alpha$ with the positive $X$-axis. The normal is perpendicular to the line,so the angle of the normal is $\alpha = 120^{\circ} - 90^{\circ} = 30^{\circ}$.The length of the perpendicular from the origin to the line is given as $p = 4$.The normal form of the equation of a line is $x \cos \alpha + y \sin \alpha = p$.Substituting the values,we get $x \cos(30^{\circ}) + y \sin(30^{\circ}) = 4$.$x(\frac{\sqrt{3}}{2}) + y(\frac{1}{2}) = 4$.Multiplying by $2$,we get $\sqrt{3}x + y = 8$.

Suppose a line makes an angle of $120^{\circ}$ with the positive direction of $X$-axis. If the length of the perpendicular from the origin to that line is $4$,then the equation of the line is

Similar Questions

If a straight line passes through the point $(-5, 4)$ and makes an intercept of length $\frac{2}{\sqrt{5}}$ between the lines $x+2y+1=0$ and $x+2y-1=0$,then the equation of that line is

If the slope of the line $ax + (3 - a)y + 7 = 0$ is $7$,then the value of the integral part of '$a$' is

If the three points are $A(2, 3)$,$B(3, 1)$,and $C(5, 3)$,then the slope of the line passing through $A$ and bisecting $BC$ is:

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

The equation of the line parallel to the line $2x - 3y = 1$ and passing through the midpoint of the line segment joining the points $(1, 3)$ and $(1, -7)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module