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(C) Given $A=(-5,-4)$ and the slope of line $L$ is $	an 	heta$. The parametric form of the line is $x = -5 + r \cos 	heta$ and $y = -4 + r \sin 	h

Vedclass · Accepted Answer

$\frac{-1}{2}$

Vedclass · Answer

(C) Given $A=(-5,-4)$ and the slope of line $L$ is $	an 	heta$. The parametric form of the line is $x = -5 + r \cos 	heta$ and $y = -4 + r \sin 	heta$. Point $B$ lies on $x+3y+2=0$,so $(-5+r_1 \cos 	heta) + 3(-4+r_1 \sin 	heta) + 2 = 0$. $-5 + r_1 \cos 	heta - 12 + 3r_1 \sin 	heta + 2 = 0 \implies r_1(\cos 	heta + 3 \sin 	heta) = 15 \implies AB = r_1 = \frac{15}{\cos 	heta + 3 \sin 	heta}$. Point $C$ lies on $2x+y+4=0$,so $2(-5+r_2 \cos 	heta) + (-4+r_2 \sin 	heta) + 4 = 0$. $-10 + 2r_2 \cos 	heta - 4 + r_2 \sin 	heta + 4 = 0 \implies r_2(2 \cos 	heta + \sin 	heta) = 10 \implies AC = r_2 = \frac{10}{2 \cos 	heta + \sin 	heta}$. Substituting into the given equation $\frac{100}{AC^2} - \frac{225}{AB^2} = 4 \cos 2	heta + \sin 2	heta$: $(2 \cos 	heta + \sin 	heta)^2 - (\cos 	heta + 3 \sin 	heta)^2 = 4 \cos 2	heta + \sin 2	heta$. $(4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta) - (\cos^2 	heta + 9 \sin^2 	heta + 6 \sin 	heta \cos 	heta) = 4(\cos^2 	heta - \sin^2 	heta) + 2 \sin 	heta \cos 	heta$. $3 \cos^2 	heta - 8 \sin^2 	heta - 2 \sin 	heta \cos 	heta = 4 \cos^2 	heta - 4 \sin^2 	heta + 2 \sin 	heta \cos 	heta$. $4 \sin^2 	heta + 4 \sin 	heta \cos 	heta + \cos^2 	heta = 0$. $(2 \sin 	heta + \cos 	heta)^2 = 0 \implies 2 \sin 	heta = -\cos 	heta \implies 	an 	heta = -\frac{1}{2}$.

List-$I$	List-$II$
$A$. The equation of line passing through $(4,3)$ whose $X$-intercept is twice its $Y$-intercept	$I$. $x+y-2\sqrt{2}=0$
$B$. The equation of the line passing through the centroid and circumcentre of $\triangle ABC$ with vertices $A(1,1), B(3,3), C(6,-6)$	$II$. $7x+23y-8=0$
$C$. The equation of the line whose $X$-intercept is $(-3/5)$ and is perpendicular to $x-y+2=0$	$III$. $x+2y+\sqrt{2}=0$
$D$. The equation of the line whose distance from the origin is $2$ and the normal drawn from the origin makes an angle $45^{\circ}$ with the positive direction of $X$-axis	$IV$. $x+2y-10=0$
	$V$. $5x+5y+3=0$

$A$ straight line $L \equiv 0$ passing through the point $A=(-5,-4)$ and having slope $\tan \theta$ meets the lines $x+3y+2=0$ and $2x+y+4=0$ respectively at the points $B$ and $C$. If $\frac{100}{AC^2}-\frac{225}{AB^2}=4 \cos 2\theta+\sin 2\theta$,then the slope of the line $L \equiv 0$ is

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