AP EAMCET · Maths · Straight Lines
The coordinate axes are rotated about the origin in the counter clockwise direction through an angle \(60^{\circ}\). If \(a\) and \(b\) are the intercepts made on the new axes by a straight line whose equation referred to the original axes is \(x+y=1\), then \(\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\)
- A \(2\)
- B \(3\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(x = x' \cos 60^{\circ} - y' \sin 60^{\circ} = \frac{x' - \sqrt{3}y'}{2}\) \(y = x' \sin 60^{\circ} + y' \cos 60^{\circ} = \frac{\sqrt{3}x' + y'}{2}\) \(\frac{x' - \sqrt{3}y'}{2} + \frac{\sqrt{3}x' + y'}{2} = 1\) \((1+\sqrt{3})x' + (1-\sqrt{3})y' = 2\)…
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