The point C(4,5) is rotated 270° counterclockwise around the origin. What are the coordinates of the resulting point, C'?
Does it show a graph or anything so I can help?
Well, C(4, 5) is all set for a little spin! Let's rotate it 270° counterclockwise around the origin.
To do that, we flip the coordinates with a little twist. Swapping x and y, and giving it some negative sign fun.
So, C'(x', y') will be (-5, 4). Ta-da! C' is the result of C getting its groove on with a 270° counterclockwise rotation!
To find the coordinates of the resulting point, C', we need to apply a rotation of 270° counterclockwise around the origin to the point C(4, 5).
To perform a rotation of a point counterclockwise around the origin, we use the following rotation formulas:
x' = x * cos(θ) - y * sin(θ)
y' = x * sin(θ) + y * cos(θ)
Where (x, y) are the original coordinates of the point, (x', y') are the coordinates of the resulting point, and θ is the angle of rotation.
Let's substitute the values into the formulas:
x' = 4 * cos(270°) - 5 * sin(270°)
y' = 4 * sin(270°) + 5 * cos(270°)
Now, we need to evaluate the trigonometric functions cosine and sine of 270°.
cos(270°) = 0
sin(270°) = -1
Substituting these values into the formulas:
x' = 4 * 0 - 5 * (-1) = 5
y' = 4 * (-1) + 5 * 0 = -4
Therefore, the coordinates of the resulting point, C', are (5, -4).
Your rotation vector will be
cos270 , - sin270
sin 270, cos 270
so multiply this rotation vector by the vector
4
5
we know cos270 = 0 , sin 270= -1
so our result is the vector
0(4) - (-1)(5) = 5
-1(4) + (0)(5) = -4
so A(4,5) ----> A'(5,-4) after a rotation of 270°
Illustrate it with a sketch, we could just as well have rotated -90°,
that might have been easier to see in our sketch.
I can also verify my result using slopes
slope OA = (5-0)/(4-0) = 5/4
slope of OA' = (-4-0)/(5-0) = -4/5
so OA and OA' form a 90° angle, or an exterior angle of 270°