JEE Mains · Maths · STD 11 - 9. straight line
Let two points be \(\mathrm{A}(1,-1)\) and \(\mathrm{B}(0,2) .\) If a point \(\mathrm{P}\left(\mathrm{x}^{\prime}, \mathrm{y}^{\prime}\right)\) be such that the area of \(\Delta \mathrm{PAB}=5\; \mathrm{sq}\) units and it lies on the line, \(3 x+y-4 \lambda=0\) then a value of \(\lambda\) is
- A \(1\)
- B \(4\)
- C \(3\)
- D \(-3\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(\overline{A B}: 3 x+y-2=0\) Also, \(\frac{1}{2} \times \sqrt{10} \times h=5\) \(\Rightarrow h=\sqrt{10}\) \(\Rightarrow \frac{|4 \lambda-2|}{\sqrt{10}}=\sqrt{10} \Rightarrow \lambda=3,-2\)
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