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
Answers
Answer:
There are two possible solutions:
Clockwise rotation
[tex]P'(x,y) = (-10,5)[/tex]
Counterclockwise rotation
[tex]P'(x,y) = (10, -5)[/tex]
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:
[tex]P'(x,y) = O(x,y) + r_{OP}\cdot (\cos (\theta_{OP}\pm \theta'),\sin (\theta_{OP}\pm \theta'))[/tex] (1)
Where:
[tex]O(x,y)[/tex] - Point of reference.
[tex]r_{OP}[/tex] - Length of the segment OP.
[tex]\theta_{OP}[/tex] - Direction of segment OP, measured in sexagesimal degrees.
[tex]\theta '[/tex] - Angle of rotation, measured in sexagesimal degrees.
Please notice that clockiwise rotation occurs when [tex]\theta = \theta_{OP}-\theta'[/tex] and counterclockwise rotation when [tex]\theta = \theta_{OP}+\theta'[/tex]. In addition, we define length and direction of the segment below:
[tex]r_{OP} = \sqrt{(x_{P}-x_{O})^{2}+(y_{P}-y_{O})^{2}}[/tex] (1)
[tex]\theta_{OP} = \tan^{-1} \frac{y_{P}-y_{O}}{x_{P}-x_{O}}[/tex]
If we know that [tex]x_{O} = y_{O} = 0[/tex], [tex]x_{P} = 5[/tex], [tex]y_{P} = 10[/tex] and [tex]\theta' = 270^{\circ}[/tex], then the coordinates of the first car after rotation is:
[tex]r_{OP} = \sqrt{(5-0)^{2}+(10-0)^{2}}[/tex]
[tex]r_{OP} \approx 11.180[/tex]
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:
[tex]\theta_{OP} = \tan^{-1} \frac{10-0}{5-0}[/tex]
[tex]\theta_{OP} \approx 63.435^{\circ}[/tex]
The two solutions are finally presented:
Clockwise rotation
[tex]P'(x,y) = (0,0) + 11.180\cdot (\cos (-206.565^{\circ}),\sin (-206.565^{\circ}))[/tex]
[tex]P'(x,y) = (-10,5)[/tex]
Counterclockwise rotation
[tex]P'(x,y) = (0,0) + 11.180\cdot (\cos (333.435^{\circ}),\sin (333.435^{\circ}))[/tex]
[tex]P'(x,y) = (10, -5)[/tex]
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