Graph the polygon with vertices A(−3,−1), B(2, 2), C(3,−3)and its image after a rotation 90 degrees about the origin.
What are the coordinates for polygon ABC after its rotation of 90 degrees about the origin?
Please show all the work on how you got your answer.
Wrong
Alright, let's get ready to spin polygon ABC around the origin!
To find the coordinates of the rotated polygon, we can use the rotation formula. For a 90-degree rotation about the origin, the formula is:
(x', y') = (xcosθ - ysinθ, xsinθ + ycosθ)
Where (x, y) represents the original coordinates, (x', y') represents the new coordinates after rotation, and θ is the angle of rotation (90 degrees in this case).
Let's apply this formula to each vertex of the polygon:
For vertex A(-3, -1):
x' = -3 * cos(90°) - (-1) * sin(90°) = 0
y' = -3 * sin(90°) + (-1) * cos(90°) = -3
So the new coordinates for A after a 90-degree rotation are (0, -3).
For vertex B(2, 2):
x' = 2 * cos(90°) - 2 * sin(90°) = -2
y' = 2 * sin(90°) + 2 * cos(90°) = 2
The new coordinates for B after rotation are (-2, 2).
For vertex C(3, -3):
x' = 3 * cos(90°) - (-3) * sin(90°) = 3
y' = 3 * sin(90°) + (-3) * cos(90°) = 3
Thus, the new coordinates for C after rotation are (3, 3).
So, polygon ABC after its rotation of 90 degrees about the origin has vertices A'(0, -3), B'(-2, 2), C'(3, 3).
I hope that wasn't too much of a twist!
To find the coordinates of the polygon after a rotation of 90 degrees about the origin, you can use the following rotational transformation formulas:
For a point (x, y) rotated counterclockwise by θ degrees about the origin, the new coordinates (x', y') can be found using the following formulas:
x' = x · cos(θ) - y · sin(θ)
y' = x · sin(θ) + y · cos(θ)
In this case, we want to rotate the polygon ABC by 90 degrees counterclockwise about the origin. Let's calculate the new coordinates for each vertex.
Vertex A(-3, -1):
x' = -3 · cos(90) - (-1) · sin(90) = -3 · 0 - (-1) · 1 = 1
y' = -3 · sin(90) + (-1) · cos(90) = -3 · 1 + (-1) · 0 = -3
Therefore, the new coordinates for vertex A after the rotation are (1, -3).
Vertex B(2, 2):
x' = 2 · cos(90) - 2 · sin(90) = 2 · 0 - 2 · 1 = -2
y' = 2 · sin(90) + 2 · cos(90) = 2 · 1 + 2 · 0 = 2
Therefore, the new coordinates for vertex B after the rotation are (-2, 2).
Vertex C(3, -3):
x' = 3 · cos(90) - (-3) · sin(90) = 3 · 0 - (-3) · 1 = 3
y' = 3 · sin(90) + (-3) · cos(90) = 3 · 1 + (-3) · 0 = 3
Therefore, the new coordinates for vertex C after the rotation are (3, 3).
Now, we can plot the original polygon ABC and its image after the rotation:
Original polygon ABC:
A(-3, -1), B(2, 2), C(3, -3)
Polygon ABC after a rotation of 90 degrees about the origin:
A'(1, -3), B'(-2, 2), C'(3, 3)
The new coordinates of the polygon after the rotation are A'(1, -3), B'(-2, 2), C'(3, 3).
Let me know if these past THREE responses to your transformtion questions helped : )
There is a song : )
It goes...
"90 degrees is just a breeze when you map negative y comma x"
that is
(x,y) becomes (-y,x)
So point A(-3,-1) has an x value of -3 and a y value of -1
So in the (-y, x)
the y value of -1 goes to the first place in the brackets, but we need -y so that is
- - 1 or 1
(1, x)
and recall our x value was -3 so (-y,x) becomes (1, -3)
Now you do the rest : )