MHT CET · Maths · Straight Lines
The line \(L\) given by \(\frac{x}{5}+\frac{y}{b} \doteq 1\) passes through the point \((13,32)\). The line K is parallel to line L and has the equation \(\frac{x}{c}+\frac{y}{3}=1\). Then the distance between L and K is \(\qquad\) units.
- A \(\frac{23}{15}\)
- 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 ).
\(\begin{array}{rr}
\therefore \quad & \frac{13}{5}+\frac{32}{b}=1 \\
& \Rightarrow b=-20
\end{array}\)
So, equation of \(L\) is \(\frac{x}{5}-\frac{y}{20}=1 \Rightarrow 4 x-y=20\)
Slope of \(L\) is \(m_1=4\).
Slope of \(\frac{x}{\mathrm{c}}+\frac{y}{3}=1\) is \(\mathrm{m}_2=\frac{3}{\mathrm{c}}\)
\(\begin{aligned}
& \Rightarrow-\frac{3}{c}=4 \\
& \Rightarrow c=\frac{3}{4}
\end{aligned}\)
Equation of line K is \(-\frac{4 x}{3}+\frac{y}{3}=1\)
\(\Rightarrow 4 x-y=-3\)
Distance between \(L\) and \(K\) is \(\left|\frac{20+3}{\sqrt{16+1}}\right|=\frac{23}{\sqrt{17}}\)
\(\begin{array}{rr}
\therefore \quad & \frac{13}{5}+\frac{32}{b}=1 \\
& \Rightarrow b=-20
\end{array}\)
So, equation of \(L\) is \(\frac{x}{5}-\frac{y}{20}=1 \Rightarrow 4 x-y=20\)
Slope of \(L\) is \(m_1=4\).
Slope of \(\frac{x}{\mathrm{c}}+\frac{y}{3}=1\) is \(\mathrm{m}_2=\frac{3}{\mathrm{c}}\)
\(\begin{aligned}
& \Rightarrow-\frac{3}{c}=4 \\
& \Rightarrow c=\frac{3}{4}
\end{aligned}\)
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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