Q101 / 180 · COMEDK 2024 (Evening Shift) · Mathematics
The line joining two points $A(2,0) B(3,1)$ is rotated about $A$ in anticlockwise direction through an angle of $15^{\circ}$. If $B$ goes to $C$ in the new position, then the coordinates of $C$ is
- A$\left(2+\dfrac{1}{\sqrt{3}}, 1\right)$
- B$\left(2+\dfrac{1}{\sqrt{3}}, \sqrt{\dfrac{3}{2}}\right)$
- Correct answerC$\left(2+\dfrac{1}{\sqrt{2}}, \sqrt{\dfrac{3}{2}}\right)$
- D$\left(2, \sqrt{\dfrac{3}{2}}\right)$
Solution
The coordinates of point $A$ are $(2, 0)$ and $B$ are $(3, 1)$. The vector $\overrightarrow{AB}$ is given by $(3-2, 1-0) = (1, 1)$.
The length of $AB$ is $r = \sqrt{1^2 + 1^2} = \sqrt{2}$. The angle $\theta$ that $AB$ makes with the positive x-axis is $\tan^{-1}(\dfrac{1}{1}) = 45^{\circ}$.
When the line is rotated anticlockwise about $A$ by $15^{\circ}$, the new angle of the line $AC$ with the positive x-axis becomes $45^{\circ} + 15^{\circ} = 60^{\circ}$.
The coordinates of $C$ are given by $(x_A + r \cos(60^{\circ}), y_A + r \sin(60^{\circ}))$.
Substituting the values, $x_C = 2 + \sqrt{2} \cos(60^{\circ}) = 2 + \sqrt{2} \times \dfrac{1}{2} = 2 + \dfrac{1}{\sqrt{2}}$.
$y_C = 0 + \sqrt{2} \sin(60^{\circ}) = \sqrt{2} \times \dfrac{\sqrt{3}}{2} = \sqrt{\dfrac{3}{2}}$.
Thus, the coordinates of $C$ are $\left(2 + \dfrac{1}{\sqrt{2}}, \sqrt{\dfrac{3}{2}}\right)$.
Answer: $\left(2+\dfrac{1}{\sqrt{2}}, \sqrt{\dfrac{3}{2}}\right)$
The length of $AB$ is $r = \sqrt{1^2 + 1^2} = \sqrt{2}$. The angle $\theta$ that $AB$ makes with the positive x-axis is $\tan^{-1}(\dfrac{1}{1}) = 45^{\circ}$.
When the line is rotated anticlockwise about $A$ by $15^{\circ}$, the new angle of the line $AC$ with the positive x-axis becomes $45^{\circ} + 15^{\circ} = 60^{\circ}$.
The coordinates of $C$ are given by $(x_A + r \cos(60^{\circ}), y_A + r \sin(60^{\circ}))$.
Substituting the values, $x_C = 2 + \sqrt{2} \cos(60^{\circ}) = 2 + \sqrt{2} \times \dfrac{1}{2} = 2 + \dfrac{1}{\sqrt{2}}$.
$y_C = 0 + \sqrt{2} \sin(60^{\circ}) = \sqrt{2} \times \dfrac{\sqrt{3}}{2} = \sqrt{\dfrac{3}{2}}$.
Thus, the coordinates of $C$ are $\left(2 + \dfrac{1}{\sqrt{2}}, \sqrt{\dfrac{3}{2}}\right)$.
Answer: $\left(2+\dfrac{1}{\sqrt{2}}, \sqrt{\dfrac{3}{2}}\right)$
Hence, the correct answer is option: 'C'.