First slide

Cartesian plane

Question

Let P be the image of the point (–3, 2) with respect to x-axis. Keeping the origin as same, the coordinate axes are rotated through an angle 60° in the clockwise sense. The coordinates of point P with respect to the new axes are

Moderate

A

$(\frac{2 \sqrt{3} - 3}{2}, - \frac{3 \sqrt{3} + 2}{2})$
B

$(\frac{2 \sqrt{3} - 3}{2}, \frac{3 \sqrt{3} + 2}{2})$
C

$(- \frac{2 \sqrt{3} - 3}{2}, \frac{3 \sqrt{3} + 2}{2})$
D

none of these

Solution

Since P is the image of the point (–3, 2) with respect to x-axis, therefore, the coordinates of P are (–3, –2).
Let (x′, y′ ) be the coordinates of P with respect to new axes. Then,
$x' = x \cos α + y \sin α = - 3 \cos (- 60^{0}) - 2 \sin (- 60^{0})$
$= - \frac{3}{2} + \sqrt{3} = \frac{2 \sqrt{3} - 3}{2}$
$y' = x \sin α + y \cos α = 3 \sin (- 60^{0}) - 2 \cos (- 60^{0})$
$= (\frac{3 \sqrt{3} + 2}{2})$
$∴$ Coordinates of P are $[\frac{2 \sqrt{3} - 3}{2}, - (\frac{3 \sqrt{3} + 2}{2})]$

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