JEE Mains · Maths · STD 11 - 9. straight line
Consider the triangles with vertices \(A (2,1) B (0,0)\) and \(C ( t , 4), t \in[0,4]\). It the maximum and the minimum perimeters of such triangles are obtained at \(t=\alpha\) and \(t=\beta\) respectively, then \(6 \alpha+21 \beta\) is equal to \(.........\).
- A \(48\)
- B \(47\)
- C \(46\)
- D \(45\)
Answer & Solution
Correct Answer
(A) \(48\)
Step-by-step Solution
Detailed explanation
\(A (2,1), B (0,0), C ( t , 4): t \in[0,4]\) \(B _1(0,8) \equiv \text { image of } B \text { w.r.t. } y=4\) for \(AC + BC + AB\) to be minimum. \(m _{ AB ^{\prime}}=\frac{-7}{2}\) \(\text { line } AB _1 \equiv 7 x +2 y =16\) \(C\left(\frac{8}{7}, 4\right)\) \(\beta=\frac{8}{7}\)…
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