Question

If line segment EF is rotated clockwise 270° about the origin, what are the coordinates of E' and F?
E (-9,-5)
F (1,6)
Explain how to

Answer 1

To rotate a point (x, y) by 270° clockwise about the origin, you can use the following formulas:

x' = x * cos(270°) - y * sin(270°)
y' = x * sin(270°) + y * cos(270°)

First, convert 270° to radians: 270° * π/180 = 3π/2

Now, apply the formulas to each point:

For point E (-9, -5):
x' = -9 * cos(3π/2) - (-5) * sin(3π/2) = -9 * 0 - (-5) * (-1) = 0 + 5 = 5
y' = -9 * sin(3π/2) + (-5) * cos(3π/2) = -9 * (-1) + (-5) * 0 = 9 + 0 = 9
Therefore, E' is at (5, 9).

For point F (1, 6):
x' = 1 * cos(3π/2) - 6 * sin(3π/2) = 1 * 0 - 6 * (-1) = 0 + 6 = 6
y' = 1 * sin(3π/2) + 6 * cos(3π/2) = 1 * (-1) + 6 * 0 = -1 + 0 = -1
Therefore, F' is at (6, -1).