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Let $d$ be the distance between the parallel lines $3 x-2 y+5=0$ and $3 x-2 y+5+2 \sqrt{13}=0$.
Let $\mathrm{L}_1 \equiv 3 \mathrm{x}-2 \mathrm{y}+\mathrm{k}_1=0\left(\mathrm{k}_1>0\right)$ and $\mathrm{L}_2 \equiv 3 \mathrm{x}-2 \mathrm{y}$ $+\mathrm{k}_2=0\left(\mathrm{k}_2>0\right)$ be two lines that are at the distance of $\frac{4 \mathrm{~d}}{\sqrt{13}}$ and $\frac{3 \mathrm{~d}}{\sqrt{13}}$ from the line $3 \mathrm{x}-2 \mathrm{y}+5=0$. Then the combined equation of the lines $\mathrm{L}_1=0$ and $\mathrm{L}_2=0$ is
Options:
Let $\mathrm{L}_1 \equiv 3 \mathrm{x}-2 \mathrm{y}+\mathrm{k}_1=0\left(\mathrm{k}_1>0\right)$ and $\mathrm{L}_2 \equiv 3 \mathrm{x}-2 \mathrm{y}$ $+\mathrm{k}_2=0\left(\mathrm{k}_2>0\right)$ be two lines that are at the distance of $\frac{4 \mathrm{~d}}{\sqrt{13}}$ and $\frac{3 \mathrm{~d}}{\sqrt{13}}$ from the line $3 \mathrm{x}-2 \mathrm{y}+5=0$. Then the combined equation of the lines $\mathrm{L}_1=0$ and $\mathrm{L}_2=0$ is
Solution:
1183 Upvotes
Verified Answer
The correct answer is:
$(3 x-2 y)^2+24(3 x-2 y)+143=0$
Distance between $3 x-2 y+5=0$ and
$$
\begin{aligned}
& 3 x-2 y+5+2 \sqrt{13}=0 \\
& d=\frac{5+2 \sqrt{13}-5}{\sqrt{3^2+2^2}}=\frac{2 \sqrt{13}}{\sqrt{13}}=2
\end{aligned}
$$
$\mathrm{L}_1: 3 x-2 y+\mathrm{k}_1 ; \mathrm{L}_2: 3 x-2 y+\mathrm{k}_2$ L: $3 x-2 y+5$
Distance between $\mathrm{L}_1 \& \mathrm{~L}=\frac{4 d}{\sqrt{13}}=\frac{8}{\sqrt{13}}$
$$
\begin{aligned}
& \frac{\left|k_1-5\right|}{\sqrt{13}}=\frac{8}{\sqrt{13}} \\
& \left|\mathrm{k}_1-5\right|=8 \Rightarrow \mathrm{k}_1=13
\end{aligned}
$$
Distance between $\mathrm{L}_2$ and $\mathrm{L}=\frac{3 d}{\sqrt{13}}=\frac{6}{\sqrt{13}}$
$$
\begin{aligned}
& \frac{\left|k_2-5\right|}{\sqrt{13}}=\frac{6}{\sqrt{13}} \\
& \left|k_2-5\right|=6 \\
& \therefore k_2=11 \therefore \mathrm{L}_1 \equiv 3 x-2 y+13=0 \\
& \mathrm{~L}_2 \equiv 3 x-2 y+11=0
\end{aligned}
$$
Combined equation of $\mathrm{L}_1$ and $\mathrm{L}_2$
$\begin{aligned} & =(3 x-2 y+13)(3 x-2 y+11)=0 \\ & \Rightarrow(3 x-2 y)^2+24(3 x-2 y)+143=0\end{aligned}$
$$
\begin{aligned}
& 3 x-2 y+5+2 \sqrt{13}=0 \\
& d=\frac{5+2 \sqrt{13}-5}{\sqrt{3^2+2^2}}=\frac{2 \sqrt{13}}{\sqrt{13}}=2
\end{aligned}
$$
$\mathrm{L}_1: 3 x-2 y+\mathrm{k}_1 ; \mathrm{L}_2: 3 x-2 y+\mathrm{k}_2$ L: $3 x-2 y+5$
Distance between $\mathrm{L}_1 \& \mathrm{~L}=\frac{4 d}{\sqrt{13}}=\frac{8}{\sqrt{13}}$
$$
\begin{aligned}
& \frac{\left|k_1-5\right|}{\sqrt{13}}=\frac{8}{\sqrt{13}} \\
& \left|\mathrm{k}_1-5\right|=8 \Rightarrow \mathrm{k}_1=13
\end{aligned}
$$
Distance between $\mathrm{L}_2$ and $\mathrm{L}=\frac{3 d}{\sqrt{13}}=\frac{6}{\sqrt{13}}$
$$
\begin{aligned}
& \frac{\left|k_2-5\right|}{\sqrt{13}}=\frac{6}{\sqrt{13}} \\
& \left|k_2-5\right|=6 \\
& \therefore k_2=11 \therefore \mathrm{L}_1 \equiv 3 x-2 y+13=0 \\
& \mathrm{~L}_2 \equiv 3 x-2 y+11=0
\end{aligned}
$$
Combined equation of $\mathrm{L}_1$ and $\mathrm{L}_2$
$\begin{aligned} & =(3 x-2 y+13)(3 x-2 y+11)=0 \\ & \Rightarrow(3 x-2 y)^2+24(3 x-2 y)+143=0\end{aligned}$
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