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

There are two possible solutions:

Clockwise rotation

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

Counterclockwise rotation

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

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) (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) (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)`