Straight Line and Pair of Straight Lines 1 Question 55
55. A line through $A(-5,-4)$ meets the line $x+3 y+2=0$, $2 x+y+4=0$ and $x-y-5=0$ at the points $B, C$ and $D$ respectively. If $(15 / A B)^{2}+(10 / A C)^{2}=(6 / A D)^{2}$, find the equation of the line.
(1993, 5M)
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Solution:
- Let equation of line $A C$ is
$$ \frac{y+4}{\sin \theta}=\frac{x+5}{\cos \theta}=r $$
Let line $A E$ make angle $\theta$ with $X$-axis and intersects $x+3 y+2=0$ at $B$ at a distance $r _1$ and line $2 x+y+4=0$ at $C$ at a distance $r _2$ and line $x-y-5=0$ at $D$ at a distance $r _3$.
$$ \begin{aligned} & \therefore \quad A B=r _1, A C=r _2, A D=r _3 . \\ & r _1=-\frac{-5-3 \times 4+2}{1 \cdot \cos \theta+3 \cdot \sin \theta} \quad \because r=-\frac{I}{(a \cos \theta+b \sin \theta)} \\ & \Rightarrow \quad r _1=\frac{15}{\cos \theta+3 \sin \theta} \\ & \Rightarrow \quad r _2=\frac{10}{2 \cos \theta+\sin \theta} \\ & \text { and } \quad r _3=-\frac{-5 \times 1-4(-1)-5}{\cos \theta-\sin \theta} \\ & \Rightarrow \quad r _3=\frac{6}{\cos \theta-\sin \theta} \end{aligned} $$
But it is given that,
$$ \begin{aligned} & \frac{15}{A B}^{2}+\frac{10}{A C}^{2}=\frac{6}{A D}^{2} \\ & \Rightarrow \quad{\frac{15}{r _1}}^{2}+{\frac{10}{r _2}}^{2}={\frac{6}{r _3}}^{2} \end{aligned} $$
$$ \begin{array}{cc} \Rightarrow & (\cos \theta+3 \sin \theta)^{2}+(2 \cos +\sin \theta)^{2}=(\cos \theta-\sin \theta)^{2} \\ & \text { [from Eqs. (i), (ii) and (iii)] } \\ \Rightarrow & \cos ^{2} \theta+9 \sin ^{2} \theta+6 \cos \theta \sin \theta+4 \cos ^{2} \theta \\ & +\sin ^{2} \theta+4 \cos \theta \sin \theta=\cos ^{2} \theta+\sin ^{2} \theta-2 \cos \theta \sin \theta \\ \Rightarrow & 4 \cos ^{2} \theta+9 \sin ^{2} \theta+12 \sin \theta \cos \theta=0 \\ \Rightarrow & (2 \cos \theta+3 \sin \theta)^{2}=0 \\ \Rightarrow & 2 \cos \theta+3 \sin \theta=0 \\ \Rightarrow & \cos \theta=-(3 / 2) \sin \theta \end{array} $$
On substituting this in equation of $A C$, we get
$$ \begin{array}{rlrl} & \frac{y+4}{\sin \theta} & =\frac{x+5}{-\frac{3}{2} \sin \theta} \\ \Rightarrow & & -3(y+4) & =2(x+5) \\ \Rightarrow & -3 y-12 & =2 x+10 \\ \Rightarrow & 2 x+3 y+22 & =0 \end{array} $$
which is the equation of required straight line.
