Question:
Let two points be \(A(1,-1)\) and \(B(0,2)\). If a point \(P(x',y')\) be such that the area of \(\Delta PAB = 5\) sq. units and it lies on the line, \(3x+y-4\lambda=0\), then a value of \(\lambda\) is :
Let two points be \(A(1,-1)\) and \(B(0,2)\). If a point \(P(x',y')\) be such that the area of \(\Delta PAB = 5\) sq. units and it lies on the line, \(3x+y-4\lambda=0\), then a value of \(\lambda\) is :
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Find the distance of P from line AB using area = (1/2) x AB x height, then put P on the given line.
Updated On: Nov 5, 2026
- \(4\)
- \(1\)
- \(-3\)
- \(3\)
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The Correct Option is D
Solution and Explanation
Line AB through \((1,-1)\) and \((0,2)\): slope \(=\frac{2-(-1)}{0-1}=-3\), so \(y=2-3x\), i.e. \(3x+y-2=0\). Also \(AB=\sqrt{1+9}=\sqrt{10}\).
Area \(=\frac12\cdot AB\cdot h = 5 \Rightarrow h=\frac{10}{\sqrt{10}}=\sqrt{10}\).
Distance of \(P(x',y')\) from AB: \(\frac{|3x'+y'-2|}{\sqrt{10}}=\sqrt{10}\Rightarrow |3x'+y'-2|=10\).
P lies on \(3x+y=4\lambda\), so \(3x'+y'=4\lambda\) and \(|4\lambda-2|=10\).
\(4\lambda-2=10\Rightarrow\lambda=3\); \(4\lambda-2=-10\Rightarrow\lambda=-2\).
Among the options only \(\lambda=3\) is available.
Answer: (4) \(3\)
Area \(=\frac12\cdot AB\cdot h = 5 \Rightarrow h=\frac{10}{\sqrt{10}}=\sqrt{10}\).
Distance of \(P(x',y')\) from AB: \(\frac{|3x'+y'-2|}{\sqrt{10}}=\sqrt{10}\Rightarrow |3x'+y'-2|=10\).
P lies on \(3x+y=4\lambda\), so \(3x'+y'=4\lambda\) and \(|4\lambda-2|=10\).
\(4\lambda-2=10\Rightarrow\lambda=3\); \(4\lambda-2=-10\Rightarrow\lambda=-2\).
Among the options only \(\lambda=3\) is available.
Answer: (4) \(3\)
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