The coordinates of the vertices of quadrilateral HIJK are H(1, 4), I(3, 2), J(−1, −4), and K(−3, −2). If quadrilateral HIJK is rotated 270° about the origin, what are the vertices of the resulting image, quadrilateral H’I’J’K’?
To find the vertices of the resulting image after rotating quadrilateral HIJK 270° about the origin, we can use the following rotation formula:
x' = x * cos(theta) - y * sin(theta)
y' = x * sin(theta) + y * cos(theta)
Where (x, y) are the coordinates of the original point, (x', y') are the coordinates of the resulting point after rotation, and theta is the angle of rotation (in this case, 270°).
Let's apply this formula to each of the vertices of quadrilateral HIJK:
For H(1, 4):
x' = 1 * cos(270°) - 4 * sin(270°) = 1 * 0 - 4 * (-1) = 4
y' = 1 * sin(270°) + 4 * cos(270°) = 1 * (-1) + 4 * 0 = -1
H'(4, -1)
For I(3, 2):
x' = 3 * cos(270°) - 2 * sin(270°) = 3 * 0 - 2 * (-1) = 2
y' = 3 * sin(270°) + 2 * cos(270°) = 3 * (-1) + 2 * 0 = -3
I'(2, -3)
For J(-1, -4):
x' = -1 * cos(270°) - (-4) * sin(270°) = -1 * 0 + 4 * (-1) = 4
y' = -1 * sin(270°) + 4 * cos(270°) = -1 * (-1) + 4 * 0 = 1
J'(4, 1)
For K(-3, -2):
x' = -3 * cos(270°) - (-2) * sin(270°) = -3 * 0 + 2 * (-1) = 2
y' = -3 * sin(270°) + 2 * cos(270°) = -3 * (-1) + 2 * 0 = 3
K'(2, 3)
Therefore, the vertices of the resulting image, quadrilateral H'I'J'K', after rotating 270° about the origin are H'(4, -1), I'(2, -3), J'(4, 1), and K'(2, 3).