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rectangular cartensian co-ordinates — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSrectangular cartensian co-ordinates1 Mark
MCQ
If $A(\cos \alpha ,\sin \alpha ),\;B(\sin \alpha , - \cos \alpha ),\,C(1,{\rm{ }}2)$ are the vertices of a $\Delta ABC$, then as $\alpha $ varies, the locus of its centroid is
  • A
    ${x^2} + {y^2} - 2x - 4y + 1 = 0$
  • ✓
    $3({x^2} + {y^2}) - 2x - 4y + 1 = 0$
  • C
    ${x^2} + {y^2} - 2x - 4y + 3 = 0$
  • D
    None of these
✓

Answer

Correct option: B.
$3({x^2} + {y^2}) - 2x - 4y + 1 = 0$
b
(b) Let $(h,\,\,k)$ be the centroid of the triangle, then

$h = \frac{{\cos \alpha + \sin \alpha + 1}}{3}$ and $k = \frac{{\sin \alpha - \cos \alpha + 2}}{3}$

$ \Rightarrow \,\,3h - 1 = \cos \alpha + \sin \alpha $ and $3k - 2 = \sin \alpha - \cos \alpha $

$ \Rightarrow \,\,{(3h - 1)^2} + {(3k - 2)^2} = 2$, (squaring and adding)

$ \Rightarrow \,9\,({h^2} + {k^2}) - 6h - 12k + 3 = 0$

$ \Rightarrow \,\,3\,({h^2} + {k^2}) - 2h - 4k + 1 = 0$

$\therefore $ Locus of $(h,\,\,k)$ is $3\,({x^2} + {y^2}) - 2x - 4y + 1 = 0$.

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