A line L passing through the point P(-5, -4) | vedclass

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(C) Let the line $L$ pass through $P(-5, -4)$ with slope $m = 	an 	heta$. The parametric equation of the line is $\frac{x+5}{\cos 	heta} = \frac{y+

Vedclass · Accepted Answer

$\pm \sqrt{3}$

Vedclass · Answer

(C) Let the line $L$ pass through $P(-5, -4)$ with slope $m = 	an 	heta$. The parametric equation of the line is $\frac{x+5}{\cos 	heta} = \frac{y+4}{\sin 	heta} = r$. Thus,$x = -5 + r \cos 	heta$ and $y = -4 + r \sin 	heta$. For point $Q$ on $x-y-5=0$: $(-5 + PQ \cos 	heta) - (-4 + PQ \sin 	heta) - 5 = 0$ $\Rightarrow PQ(\cos 	heta - \sin 	heta) = 6$ $\Rightarrow \frac{6}{PQ} = \cos 	heta - \sin 	heta$. Multiplying by $3$,we get $\frac{18}{PQ} = 3 \cos 	heta - 3 \sin 	heta$ ... $(i)$. For point $R$ on $x+3y+2=0$: $(-5 + PR \cos 	heta) + 3(-4 + PR \sin 	heta) + 2 = 0$ $\Rightarrow PR(\cos 	heta + 3 \sin 	heta) = 15$ $\Rightarrow \frac{15}{PR} = \cos 	heta + 3 \sin 	heta$ ... $(ii)$. Given $\frac{18}{PQ} + \frac{15}{PR} = 2$,substituting $(i)$ and $(ii)$: $(3 \cos 	heta - 3 \sin 	heta) + (\cos 	heta + 3 \sin 	heta) = 2$ $\Rightarrow 4 \cos 	heta = 2$ $\Rightarrow \cos 	heta = \frac{1}{2}$. Since $\cos 	heta = \frac{1}{2}$,$\sin 	heta = \pm \sqrt{1 - (\frac{1}{2})^2} = \pm \frac{\sqrt{3}}{2}$. Slope $m = 	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\pm \sqrt{3}/2}{1/2} = \pm \sqrt{3}$.

$A$ line $L$ passing through the point $P(-5, -4)$ cuts the lines $x-y-5=0$ and $x+3y+2=0$ at $Q$ and $R$ respectively such that $\frac{18}{PQ} + \frac{15}{PR} = 2$. Then the slope of the line $L$ is:

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