Question:
Let \(L\) be the line passing through the point \(P(1,2)\) such that its intercepted segment between the co-ordinate axes is bisected at \(P\). If \(L_1\) is the line perpendicular to \(L\) and passing through the point \((-2,\,1)\), then the point of intersection of \(L\) and \(L_1\) is
Let \(L\) be the line passing through the point \(P(1,2)\) such that its intercepted segment between the co-ordinate axes is bisected at \(P\). If \(L_1\) is the line perpendicular to \(L\) and passing through the point \((-2,\,1)\), then the point of intersection of \(L\) and \(L_1\) is
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The midpoint condition gives intercepts 2 and 4, so \(L: 2x+y=4\); then intersect with the perpendicular through \((-2,1)\).
Updated On: Nov 11, 2026
- \(\left(\frac{3}{5},\frac{23}{10}\right)\)
- \(\left(\frac{4}{5},\frac{12}{5}\right)\)
- \(\left(\frac{11}{20},\frac{29}{10}\right)\)
- \(\left(\frac{3}{10},\frac{17}{5}\right)\)
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The Correct Option is B
Solution and Explanation
Let \(L\) meet the axes at \((a,0)\) and \((0,b)\). Midpoint \(=(\frac a2,\frac b2)=(1,2)\Rightarrow a=2,\ b=4\).
\(L:\ \frac x2+\frac y4=1\Rightarrow 2x+y=4\), slope \(-2\).
\(L_1\) is perpendicular with slope \(\frac12\) through \((-2,1)\): \(y-1=\frac12(x+2)\Rightarrow x-2y+4=0\).
Substitute \(y=4-2x\): \(x-8+4x+4=0\Rightarrow x=\frac45,\ y=\frac{12}5\).
Answer: (2) \(\left(\frac{4}{5},\frac{12}{5}\right)\)
\(L:\ \frac x2+\frac y4=1\Rightarrow 2x+y=4\), slope \(-2\).
\(L_1\) is perpendicular with slope \(\frac12\) through \((-2,1)\): \(y-1=\frac12(x+2)\Rightarrow x-2y+4=0\).
Substitute \(y=4-2x\): \(x-8+4x+4=0\Rightarrow x=\frac45,\ y=\frac{12}5\).
Answer: (2) \(\left(\frac{4}{5},\frac{12}{5}\right)\)
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