AP EAMCET · Maths · Straight Lines
A point \(P\) on a line is at a distance of 4 units from the origin \((0,0)\). If the line makes \(60^{\circ}\) with the negative direction of the \(X\)-axis, then \(P\) is
- A \((2,2 \sqrt{3})\)
- B \((2 \sqrt{3}, 2)\)
- C \((1, \sqrt{3})\)
- D \((2 \sqrt{3}, 1)\)
Answer & Solution
Correct Answer
(B) \((2 \sqrt{3}, 2)\)
Step-by-step Solution
Detailed explanation
\(O P\) is perpendicular to given line. \(\therefore\) Slope of \(O P=\frac{1}{\sqrt{3}}=\frac{y}{x} \Rightarrow x=\sqrt{3} y\) Also \(x^2+y^2=16\) \(\therefore 3 y^2+y^2=16 \Rightarrow y= \pm 2\) \(\therefore\) Points can be \((2 \sqrt{3}, 2)\) or \((-2 \sqrt{3},-2)\).
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