A line L through A(-5,-4) meets the lines x+3 | vedclass

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Question

(A) Let the equation of the line passing through $A(-5,-4)$ be $\frac{x+5}{\cos 	heta} = \frac{y+4}{\sin 	heta} = r$. Any point on this line is $(-5

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$2x+3y+22=0$

Vedclass · Answer

(A) Let the equation of the line passing through $A(-5,-4)$ be $\frac{x+5}{\cos 	heta} = \frac{y+4}{\sin 	heta} = r$. Any point on this line is $(-5+r\cos 	heta, -4+r\sin 	heta)$. For point $B$ on $x+3y+2=0$: $(-5+r_1\cos 	heta) + 3(-4+r_1\sin 	heta) + 2 = 0$ $\Rightarrow r_1(\cos 	heta + 3\sin 	heta) = 15$ $\Rightarrow \frac{15}{r_1} = \cos 	heta + 3\sin 	heta \dots (i)$. For point $C$ on $2x+y+4=0$: $2(-5+r_2\cos 	heta) + (-4+r_2\sin 	heta) + 4 = 0$ $\Rightarrow r_2(2\cos 	heta + \sin 	heta) = 10$ $\Rightarrow \frac{10}{r_2} = 2\cos 	heta + \sin 	heta \dots (ii)$. For point $D$ on $x-y-5=0$: $(-5+r_3\cos 	heta) - (-4+r_3\sin 	heta) - 5 = 0$ $\Rightarrow r_3(\cos 	heta - \sin 	heta) = 6$ $\Rightarrow \frac{6}{r_3} = \cos 	heta - \sin 	heta \dots (iii)$. Given $\left(\frac{15}{AB}ight)^2 + \left(\frac{10}{AC}ight)^2 = \left(\frac{6}{AD}ight)^2$,we have: $(\cos 	heta + 3\sin 	heta)^2 + (2\cos 	heta + \sin 	heta)^2 = (\cos 	heta - \sin 	heta)^2$. $(\cos^2 	heta + 9\sin^2 	heta + 6\sin 	heta \cos 	heta) + (4\cos^2 	heta + \sin^2 	heta + 4\cos 	heta \sin 	heta) = \cos^2 	heta + \sin^2 	heta - 2\sin 	heta \cos 	heta$. $5\cos^2 	heta + 10\sin^2 	heta + 10\sin 	heta \cos 	heta = \cos^2 	heta + \sin^2 	heta - 2\sin 	heta \cos 	heta$. $4\cos^2 	heta + 9\sin^2 	heta + 12\sin 	heta \cos 	heta = 0$. $(2\cos 	heta + 3\sin 	heta)^2 = 0 \Rightarrow 2\cos 	heta + 3\sin 	heta = 0$. $	an 	heta = -\frac{2}{3}$. The equation of line $L$ is $y - (-4) = -\frac{2}{3}(x - (-5))$ $\Rightarrow 3y + 12 = -2x - 10$ $\Rightarrow 2x + 3y + 22 = 0$.

$A$ line $L$ through $A(-5,-4)$ meets the lines $x+3y+2=0$,$2x+y+4=0$,and $x-y-5=0$ at points $B$,$C$,and $D$ respectively. If $\left(\frac{15}{AB}\right)^2+\left(\frac{10}{AC}\right)^2=\left(\frac{6}{AD}\right)^2$,then find the equation of $L$.

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