The equation of a line,whose perpendicular di | vedclass

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(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin

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$x \sqrt{3}+y+10=0$

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(C) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle made by the perpendicular with the positive $X$-axis.Given $p = 5$ and $\alpha = 210^{\circ}$.Substituting these values into the equation:$x \cos(210^{\circ}) + y \sin(210^{\circ}) = 5$Since $\cos(210^{\circ}) = \cos(180^{\circ} + 30^{\circ}) = -\cos(30^{\circ}) = -\frac{\sqrt{3}}{2}$ and $\sin(210^{\circ}) = \sin(180^{\circ} + 30^{\circ}) = -\sin(30^{\circ}) = -\frac{1}{2}$,$x(-\frac{\sqrt{3}}{2}) + y(-\frac{1}{2}) = 5$Multiplying both sides by $-2$:$\sqrt{3}x + y = -10$$\sqrt{3}x + y + 10 = 0$

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 equation of a line,whose perpendicular distance from the origin is $5$ units and the angle,which the perpendicular to the line from the origin makes,is $210^{\circ}$ with the positive $X$-axis,is

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