Question:
Suppose that the points \((h, k)\), \((1, 2)\) 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 \(\frac{k}{h}\) equals :
Suppose that the points \((h, k)\), \((1, 2)\) 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 \(\frac{k}{h}\) equals :
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Slope of \(L_1\) from the two given points; \(L_2\) has the negative reciprocal slope and passes through \((4, 3)\) and \((h, k)\).
Updated On: Nov 5, 2026
- \(\frac{1}{3}\)
- \(3\)
- \(0\)
- \(-\frac{1}{7}\)
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The Correct Option is A
Solution and Explanation
Slope of \(L_1\): \(\frac{4 - 2}{-3 - 1} = -\frac{1}{2}\). Since \(L_2 \perp L_1\), slope of \(L_2 = 2\).
\(L_2\) passes through \((h, k)\) and \((4, 3)\): \(\frac{k - 3}{h - 4} = 2 \Rightarrow k = 2h - 5\).
\((h, k)\) lies on \(L_1\): \(k - 2 = -\frac{1}{2}(h - 1) \Rightarrow 2k - 4 = -h + 1 \Rightarrow h + 2k = 5\).
Substituting: \(h + 4h - 10 = 5 \Rightarrow h = 3,\ k = 1\).
\(\frac{k}{h} = \frac{1}{3}\).
Answer: (1) \(\frac{1}{3}\)
\(L_2\) passes through \((h, k)\) and \((4, 3)\): \(\frac{k - 3}{h - 4} = 2 \Rightarrow k = 2h - 5\).
\((h, k)\) lies on \(L_1\): \(k - 2 = -\frac{1}{2}(h - 1) \Rightarrow 2k - 4 = -h + 1 \Rightarrow h + 2k = 5\).
Substituting: \(h + 4h - 10 = 5 \Rightarrow h = 3,\ k = 1\).
\(\frac{k}{h} = \frac{1}{3}\).
Answer: (1) \(\frac{1}{3}\)
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