MHT CET · Maths · Three Dimensional Geometry
The line L given by \(\frac{x}{5}+\frac{y}{b}=1\) passes through the point \((13,32)\). The line K is parallel to L and has the equation \(\frac{x}{c}+\frac{y}{3}=1\). Then the distance between L and K is
- A \(\frac{23}{\sqrt{15}}\)
- B \(\sqrt{17}\)
- C \(\frac{17}{\sqrt{15}}\)
- D \(\frac{23}{\sqrt{17}}\)
Answer & Solution
Correct Answer
(D) \(\frac{23}{\sqrt{17}}\)
Step-by-step Solution
Detailed explanation
Line L passes through \((13,32)\).
\(\begin{aligned}
\therefore \quad & \frac{13}{5}+\frac{32}{b}=1 \\
& \Rightarrow b=-20
\end{aligned}\)
So, equation of L is \(\frac{x}{5}-\frac{y}{20}=1 \Rightarrow 4 x-y=20\)
Slope of \(L\) is \(\mathrm{m}_1=4\).
Slope of \(\frac{x}{c}+\frac{y}{3}=1\) is \(\mathrm{m}_2=-\frac{3}{c}\)
\(\begin{aligned}
& \Rightarrow-\frac{3}{c}=4 \\
& \Rightarrow c=-\frac{3}{4}
\end{aligned}\)
Equation of line K is \(-\frac{4 x}{3}+\frac{y}{3}=1\)
\(\Rightarrow 4 x-y=-3\)
Distance between \(L\) and \(K\) is \(\left|\frac{20+3}{\sqrt{16+1}}\right|=\frac{23}{\sqrt{17}}\)
\(\begin{aligned}
\therefore \quad & \frac{13}{5}+\frac{32}{b}=1 \\
& \Rightarrow b=-20
\end{aligned}\)
So, equation of L is \(\frac{x}{5}-\frac{y}{20}=1 \Rightarrow 4 x-y=20\)
Slope of \(L\) is \(\mathrm{m}_1=4\).
Slope of \(\frac{x}{c}+\frac{y}{3}=1\) is \(\mathrm{m}_2=-\frac{3}{c}\)
\(\begin{aligned}
& \Rightarrow-\frac{3}{c}=4 \\
& \Rightarrow c=-\frac{3}{4}
\end{aligned}\)
Equation of line K is \(-\frac{4 x}{3}+\frac{y}{3}=1\)
\(\Rightarrow 4 x-y=-3\)
Distance between \(L\) and \(K\) is \(\left|\frac{20+3}{\sqrt{16+1}}\right|=\frac{23}{\sqrt{17}}\)
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