Question

A car ride is drawn on a coordinate plane so that the first card is located at the point by (5,10) what are the coordinates of the first car after a rotation of 270° about the origin

Answer 1

Answer:

There are two possible solutions:

Clockwise rotation

$P'(x,y) = (-10,5)$

Counterclockwise rotation

$P'(x,y) = (10, -5)$

Step-by-step explanation:

There are two possible answers: (i) Clockwise rotation, (ii) Counterclockwise rotation. Vectorially speaking, rotation of point of rotation of a point about another point of reference is defined by:

$P'(x,y) = O(x,y) + r_{OP}\cdot (\cos (\theta_{OP}\pm \theta'),\sin (\theta_{OP}\pm \theta'))$ (1)

Where:

$O(x,y)$ - Point of reference.

$r_{OP}$ - Length of the segment OP.

$\theta_{OP}$ - Direction of segment OP, measured in sexagesimal degrees.

$\theta '$ - Angle of rotation, measured in sexagesimal degrees.

Please notice that clockiwise rotation occurs when $\theta = \theta_{OP}-\theta'$ and counterclockwise rotation when $\theta = \theta_{OP}+\theta'$. In addition, we define length and direction of the segment below:

$r_{OP} = \sqrt{(x_{P}-x_{O})^{2}+(y_{P}-y_{O})^{2}}$ (1)

$\theta_{OP} = \tan^{-1} \frac{y_{P}-y_{O}}{x_{P}-x_{O}}$

If we know that $x_{O} = y_{O} = 0$, $x_{P} = 5$, $y_{P} = 10$ and $\theta' = 270^{\circ}$, then the coordinates of the first car after rotation is:

$r_{OP} = \sqrt{(5-0)^{2}+(10-0)^{2}}$

$r_{OP} \approx 11.180$

Please notice that original point is located at first quadrant of the Cartesian plane centered at origin, then the direction of the segment OP is:

$\theta_{OP} = \tan^{-1} \frac{10-0}{5-0}$

$\theta_{OP} \approx 63.435^{\circ}$

The two solutions are finally presented:

Clockwise rotation

$P'(x,y) = (0,0) + 11.180\cdot (\cos (-206.565^{\circ}),\sin (-206.565^{\circ}))$

$P'(x,y) = (-10,5)$

Counterclockwise rotation

$P'(x,y) = (0,0) + 11.180\cdot (\cos (333.435^{\circ}),\sin (333.435^{\circ}))$

$P'(x,y) = (10, -5)$