The line passing through (-1, pi/2) and perpe | vedclass

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Question

(A) The given line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$.Any line perpendicular to $A \sin 	heta + B \cos 	heta = \frac{C}{r}$ is

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given line is $\sqrt{3} \sin 	heta + 2 \cos 	heta = \frac{4}{r}$.Any line perpendicular to $A \sin 	heta + B \cos 	heta = \frac{C}{r}$ is of the form $A \sin(	heta + \pi/2) + B \cos(	heta + \pi/2) = \frac{k}{r}$,which simplifies to $A \cos 	heta - B \sin 	heta = \frac{k}{r}$.Substituting the values,we get $\sqrt{3} \cos 	heta - 2 \sin 	heta = \frac{k}{r}$.Since the line passes through $(-1, \pi/2)$,we substitute $r = -1$ and $	heta = \pi/2$:$\sqrt{3} \cos(\pi/2) - 2 \sin(\pi/2) = \frac{k}{-1}$.$0 - 2(1) = -k \Rightarrow k = 2$.Thus,the equation of the line is $\sqrt{3} \cos 	heta - 2 \sin 	heta = \frac{2}{r}$,which can be written as $2 = \sqrt{3} r \cos 	heta - 2 r \sin 	heta$.

List-$I$	List-$II$
$A$. Line passing through $(-4, 3)$ and having intercepts in the ratio $5:3$	$1$. $2x - 5y + 4 = 0$
$B$. Line passing through $P(2, -5)$ such that $P$ bisects the part intercepted between the axes	$2$. $3x + 5y = 3$
$C$. Line parallel to $2x - 3y + 5 = 0$ with $x$-intercept $\frac{2}{5}$ is	$3$. $10x - 15y + 4 = 0$
$D$. Line perpendicular to $5x + 2y + 7 = 0$ with $y$-intercept $\frac{4}{5}$ is	$4$. $10x - 15y = 4$
	$5$. $5x - 2y - 20 = 0$

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

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