MHT CET · Maths · Straight Lines
Suppose that the points \((\mathrm{h}, \mathrm{k}),(1,2)\) and \((-3,4)\) lie on the line \(l_1\). If a line \(l_2\) passing through the points \((h, k)\) and \((4,3)\) is perpendicular to \(l_1\), then \(\left(\frac{k}{h}\right)\) equals
- A \(\frac{1}{3}\)
- B 0
- C 3
- D \(-\frac{1}{7}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Slope of line \(l_1=\frac{4-2}{-3-1}=-\frac{1}{2}\)
Equation of \(l_1\) is
\(y-4=-\frac{1}{2}(x+3)\)
\(\Rightarrow x+2 y=5\)...(i)
Since \(l_1 \perp l_2\)
\(\therefore \quad\) Slope of \(l_2=2\)
Equation of \(l_2\) is
\(\begin{aligned}
& y-3=2(x-4) \\
& \Rightarrow 2 x-y=5...(ii)
\end{aligned}\)
Solving (i) and (ii), we get
\(\begin{aligned}
& x=3, y=1 \\
& \Rightarrow \mathrm{~h}=3, \mathrm{k}=1 \\
& \Rightarrow \frac{\mathrm{k}}{\mathrm{~h}}=\frac{1}{3}
\end{aligned}\)
Equation of \(l_1\) is
\(y-4=-\frac{1}{2}(x+3)\)
\(\Rightarrow x+2 y=5\)...(i)
Since \(l_1 \perp l_2\)
\(\therefore \quad\) Slope of \(l_2=2\)
Equation of \(l_2\) is
\(\begin{aligned}
& y-3=2(x-4) \\
& \Rightarrow 2 x-y=5...(ii)
\end{aligned}\)
Solving (i) and (ii), we get
\(\begin{aligned}
& x=3, y=1 \\
& \Rightarrow \mathrm{~h}=3, \mathrm{k}=1 \\
& \Rightarrow \frac{\mathrm{k}}{\mathrm{~h}}=\frac{1}{3}
\end{aligned}\)
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