AP EAMCET · Maths · Straight Lines
If \(\mathrm{A}(\cos \alpha, \sin \alpha), \mathrm{B}(\sin \alpha,-\cos \alpha), \mathrm{C}(1,2)\) are the vertices of a \(\triangle \mathrm{ABC}\) then the locus of its centroid is
- A \(3\left(x^2+y^2\right)-2 x-4 y+1=0\)
- B \(x^2+y^2-2 x-4 y+1=0\)
- C \(x^2+y^2-2 x-4 y+3=0\)
- D \(2\left(x^2+y^2\right)-2 x-4 y+5=0\)
Answer & Solution
Correct Answer
(A) \(3\left(x^2+y^2\right)-2 x-4 y+1=0\)
Step-by-step Solution
Detailed explanation
Let \( (x, y) \) be the coordinates of the centroid G. \( x = \frac{\cos \alpha + \sin \alpha + 1}{3} \Rightarrow 3x - 1 = \cos \alpha + \sin \alpha \) \( y = \frac{\sin \alpha - \cos \alpha + 2}{3} \Rightarrow 3y - 2 = \sin \alpha - \cos \alpha \)…
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