Find the values of theta and p,if the equatio | vedclass

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

Question

(A) The equation of the given line is $\sqrt{3} x + y + 2 = 0$. This equation can be rewritten as $-\sqrt{3} x - y = 2$. Dividing both sides by $\sqrt

Vedclass · Accepted Answer

$	heta = \frac{7 \pi}{6}, p = 1$

Vedclass · Answer

(A) The equation of the given line is $\sqrt{3} x + y + 2 = 0$. This equation can be rewritten as $-\sqrt{3} x - y = 2$. Dividing both sides by $\sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$,we get: $-\frac{\sqrt{3}}{2} x - \frac{1}{2} y = 1$. Comparing this with the normal form $x \cos 	heta + y \sin 	heta = p$,we have: $\cos 	heta = -\frac{\sqrt{3}}{2}$,$\sin 	heta = -\frac{1}{2}$,and $p = 1$. Since both $\sin 	heta$ and $\cos 	heta$ are negative,$	heta$ lies in the third quadrant. $	heta = \pi + \frac{\pi}{6} = \frac{7 \pi}{6}$. Thus,the values are $	heta = \frac{7 \pi}{6}$ and $p = 1$.

Find the values of $\theta$ and $p$,if the equation $x \cos \theta + y \sin \theta = p$ is the normal form of the line $\sqrt{3} x + y + 2 = 0$.

Similar Questions

$A$ straight line parallel to the line $y=\sqrt{3}x$ passes through $Q(2,3)$ and cuts the line $2x+4y-27=0$ at $P$. Then the length of the line segment $PQ$ is

In what direction should a line be drawn through the point $(1, 2)$ so that its point of intersection with the line $x + y = 4$ is at a distance of $\frac{\sqrt{6}}{3}$ from the given point (in $^\circ$)?

The slope of a line that makes intercepts of equal length on the axes is:

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

The polar equation of the line perpendicular to the line $\sin \theta - \cos \theta = \frac{1}{r}$ and passing through the point $\left(2, \frac{\pi}{6}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module