The line passing through left(-1, frac{pi}{2} | vedclass

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

Question

(A) Given equation of the line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin 	heta + 2 r \cos 	he

Vedclass · Accepted Answer

$2 = \sqrt{3} r \cos 	heta - 2 r \sin 	heta$

Vedclass · Answer

(A) Given equation of the line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$,which can be written as $\sqrt{3} r \sin 	heta + 2 r \cos 	heta = 4$. Converting to Cartesian coordinates using $x = r \cos 	heta$ and $y = r \sin 	heta$,we get $\sqrt{3} y + 2 x = 4$,or $2x + \sqrt{3} y - 4 = 0$. The slope of this line is $m_1 = -\frac{2}{\sqrt{3}}$. The slope of a line perpendicular to this is $m_2 = -\frac{1}{m_1} = \frac{\sqrt{3}}{2}$. The point $\left(-1, \frac{\pi}{2}ight)$ in polar coordinates corresponds to Cartesian coordinates $(x_1, y_1) = (-1 \cos \frac{\pi}{2}, -1 \sin \frac{\pi}{2}) = (0, -1)$. The equation of the line passing through $(0, -1)$ with slope $\frac{\sqrt{3}}{2}$ is $y - (-1) = \frac{\sqrt{3}}{2}(x - 0)$. $y + 1 = \frac{\sqrt{3}}{2} x$ $\Rightarrow 2y + 2 = \sqrt{3} x$ $\Rightarrow \sqrt{3} x - 2y = 2$. Substituting $x = r \cos 	heta$ and $y = r \sin 	heta$,we get $\sqrt{3} r \cos 	heta - 2 r \sin 	heta = 2$.

The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta + 2 \cos \theta = \frac{4}{r}$ is:

Similar Questions

Find the equation of the line passing through the point $(-1, 2)$ and parallel to the line $3x - 4y + 5 = 0$.

$P(6,4)$ is a point on the line $x-y-2=0$. If $A(\alpha, \beta)$ and $B(\gamma, \delta)$ are two points on this line lying on either side of $P$ at a distance of $4$ units from $P$,then $\alpha^2+\beta^2+\gamma^2+\delta^2=$

The equation of the straight line passing through $(-a, 0)$ and forming a triangle with the coordinate axes of area $T$ is

For what values of $a$ and $b$ are the intercepts cut off on the coordinate axes by the line $ax + by + 8 = 0$ equal in length but opposite in sign to those cut off by the line $2x - 3y + 6 = 0$ on the axes?

The line $MN$ whose equation is $x-y-2=0$ cuts the $X$-axis at $M$ and the coordinates of $N$ are $(4,2)$. The line $MN$ is rotated about $M$ through $45^{\circ}$ in the anticlockwise direction. What is the equation of the line $MN$ in the new position?

Vedclass Test Series

Exam Paper Generator

Online Exam Module