JEE Mains · Maths · STD 11 - 9. straight line
If the point \(\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) lies on the curve traced by the mid-points of the line segments of the lines \(x\) \(\cos \theta+ y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)\) between the coordinates axes, then \(\alpha\) is equal to
- A \(7\)
- B \(-7\)
- C \(-7 \sqrt{3}\)
- D \(7 \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(7\)
Step-by-step Solution
Detailed explanation
\(\operatorname{pt}\left(\alpha, \frac{7 \sqrt{3}}{3}\right)\) \(x -\text { intercept }=\frac{7}{\cos \theta}\) \(y -\text { intercept }=\frac{7}{\sin \theta}\) \(A:\left(\frac{7}{\cos \theta}, 0\right) B :\left(0, \frac{7}{\sin \theta}\right)\) Locus of mid pt \(M :( h , k )\)…
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