JEE Mains · Maths · STD 11 - 9. straight line
Let \(\mathrm{A}\) be the set of all points \((\alpha, \beta)\) such that the area of triangle formed by the points \((5,6),(3,2)\) and \((\alpha, \beta)\) is \(12\, square\, units.\) Then the least possible length of a line segment joining the origin to a point in \(A,\) is :
- A \(\frac{4}{\sqrt{5}}\)
- B \(\frac{16}{\sqrt{5}}\)
- C \(\frac{8}{\sqrt{5}}\)
- D \(\frac{12}{\sqrt{5}}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{\sqrt{5}}\)
Step-by-step Solution
Detailed explanation
\(\left|\frac{1}{2}\right| \begin{array}{lll}5 & 6 & 1 \\ 3 & 2 & 1 \\ \alpha & \beta & 1\end{array}||=12\) \(4 \alpha-2 \beta=\pm 24+8\) \(\Rightarrow 4 \alpha-2 \beta=+24+8 \Rightarrow 2 \alpha-\beta=16\) \(2 \mathrm{x}-\mathrm{y}-16=0...(1)\)…
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