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

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Aicharya · Accepted Answer

Let equation of line passing through $A(-5,-4)$ is given by

$ \frac{x-(-5)}{\cos \theta}=\frac{y-(-4)}{\sin \theta}=r \Rightarrow \frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}=r $ Therefore, every point on the line is of the form $ x=-5+r \cos \theta, y=-4+r \sin \theta $ Let $A B=r_1, A C=r_2, A D=r_3$ $ \begin{aligned} & \because \quad B=\left(-5+r_1 \cos \theta,-4+r_1 \sin \theta\right) \text { and lies on } \\ & x+3 y+2=0 \\ & \Rightarrow-5+r_1 \cos \theta+3\left(-4+r_1 \sin \theta\right)+2=0 \\ & \Rightarrow \quad r_1=\frac{15}{\cos \theta+3 \sin \theta} \\ & \Rightarrow \quad \frac{15}{r_1}=\cos \theta+3 \sin \theta...(i) \end{aligned} $ $\because C=\left(-5+r_2 \cos \theta,-4+r_2 \sin \theta\right)$ and lies on $2 x+y+4=0$ $ \Rightarrow 2\left(-5+r_2 \cos \theta\right)+\left(-4+r_2 \sin \theta\right)+4=0 $ $ \Rightarrow \frac{10}{r_2}=2 \cos \theta+\sin \theta...(ii) $ $\because D=\left(-5+r_3 \cos \theta,-4+r_3 \sin \theta\right)$ and lies on $x-y-5=0$ $ \begin{aligned} & \Rightarrow \quad-5+r_3 \cos \theta-\left(-4+r_3 \sin \theta\right)-5=0 \\ & \Rightarrow \quad \frac{6}{r_3}=\cos \theta-\sin \theta...(iii) \end{aligned} $ $\begin{aligned} & \text { Given that, }\left(\frac{15}{A B}\right)^2+\left(\frac{10}{A C}\right)^2=\left(\frac{6}{A P}\right)^2 \\ & \Rightarrow \quad\left(\frac{15}{r_1}\right)^2+\left(\frac{10}{r_2}\right)^2=\left(\frac{6}{r_3}\right)^2 \\ & \Rightarrow(\cos \theta+3 \sin \theta)^2+(2 \cos \theta+\sin \theta)^2 \\ & =(\cos \theta-\sin \theta)^2 \quad \text { [by Eqs. (i), (ii) and (iii)] } \\ & \Rightarrow \cos ^2 \theta+9 \sin ^2 \theta+6 \sin \theta \cos \theta \\ & +4 \cos ^2 \theta+\sin ^2 \theta+4 \cos \theta \sin \theta \\ & =\cos ^2 \theta+\sin ^2 \theta-2 \sin \theta \cos \theta \\ & \Rightarrow 4 \cos ^2 \theta+9 \sin ^2 \theta+12 \sin \theta \cos \theta=0 \\ & \Rightarrow \quad(2 \cos \theta+3 \sin \theta)^2=0 \\ & \Rightarrow \quad 2 \cos \theta+3 \sin \theta=0 \\ & \Rightarrow \quad 3 \sin \theta=-2 \cos \theta\end{aligned}$ $ \Rightarrow \quad \frac{\sin \theta}{\cos \theta}=-\frac{2}{3} \Rightarrow \tan \theta=-\frac{2}{3} $ Equation of line passing through $A(-5,-4)$ is $ \begin{aligned} & y+4=\tan \theta(x+5) \\ & \Rightarrow \quad y+4=-\frac{2}{3}(x+5) \\ & \Rightarrow \quad 3 y+12=-2 x-10 \Rightarrow 2 x+3 y+22=0 \end{aligned} $

A line $L$ through $A(-5,-4)$ meets the lines $x+3 y+2=0$ and $2 x+y+4=0$, $x-y-5=0$ at points $B, C$ and…

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