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Question

(B) Let the equation of the line passing through $A(-5, -4)$ with inclination $	heta$ be $\frac{x + 5}{\cos 	heta} = \frac{y + 4}{\sin 	heta} = r$.

Vedclass · Accepted Answer

$2x + 3y + 22 = 0$

Vedclass · Answer

(B) Let the equation of the line passing through $A(-5, -4)$ with inclination $	heta$ be $\frac{x + 5}{\cos 	heta} = \frac{y + 4}{\sin 	heta} = r$. Any point on this line is $(-5 + r \cos 	heta, -4 + r \sin 	heta)$. Since $B$,$C$,and $D$ lie on the given lines,we substitute the coordinates into the equations: For $x + 3y + 2 = 0$: $(-5 + AB \cos 	heta) + 3(-4 + AB \sin 	heta) + 2 = 0 \implies AB(\cos 	heta + 3 \sin 	heta) = 15 \implies \frac{15}{AB} = \cos 	heta + 3 \sin 	heta$. For $2x + y + 4 = 0$: $2(-5 + AC \cos 	heta) + (-4 + AC \sin 	heta) + 4 = 0 \implies AC(2 \cos 	heta + \sin 	heta) = 10 \implies \frac{10}{AC} = 2 \cos 	heta + \sin 	heta$. For $x - y - 5 = 0$: $(-5 + AD \cos 	heta) - (-4 + AD \sin 	heta) - 5 = 0 \implies AD(\cos 	heta - \sin 	heta) = 6 \implies \frac{6}{AD} = \cos 	heta - \sin 	heta$. Given the condition: $(\cos 	heta + 3 \sin 	heta)^2 + (2 \cos 	heta + \sin 	heta)^2 = (\cos 	heta - \sin 	heta)^2$. Expanding: $(\cos^2 	heta + 9 \sin^2 	heta + 6 \sin 	heta \cos 	heta) + (4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta) = \cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$. $5 \cos^2 	heta + 10 \sin^2 	heta + 10 \sin 	heta \cos 	heta = \cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta$. $4 \cos^2 	heta + 9 \sin^2 	heta + 12 \sin 	heta \cos 	heta = 0$. $(2 \cos 	heta + 3 \sin 	heta)^2 = 0 \implies 	an 	heta = -\frac{2}{3}$. The equation of the line is $y + 4 = -\frac{2}{3}(x + 5) \implies 3y + 12 = -2x - 10 \implies 2x + 3y + 22 = 0$.

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

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