Question

Rectangle CDEF with vertices C(-1,6), D(0,8), E(6,5) and F(5,3): 270 degrees counterclockwise

Answer 1

The vertices of the image are $C'(x,y) = (6, 1)$, $D'(x,y) = (8, 0)$, $E'(x,y) = (5, -6)$ and $F'(x,y) = (3, -5)$.

Given a point $(x,y)$, we define the rotation of that point with respect to origin by this operation:

$(x',y') = (x\cdot \cos \theta - y\cdot \sin \theta, x\cdot \sin \theta + y\cdot \cos \theta)$ (1)

Where a counterclockwise rotation occurs when $\theta > 0^{\circ}$.

If we know that $C(x,y) = (-1,6)$, $D(x,y) = (0, 8)$, $E(x,y) = (6,5)$, $F(x,y) = (5,3)$ and $\theta = 270^{\circ}$, then the images of such points are:

$C'(x, y) = (-1\cdot \cos 270^{\circ}-6\cdot \sin 270^{\circ}, -1\cdot \sin 270^{\circ}+6\cdot \cos 270^{\circ})$

$C'(x,y) = (6, 1)$

$D'(x,y) = (0\cdot \cos 270^{\circ}-8\cdot \sin 270^{\circ}, 0\cdot \sin 270^{\circ}+8\cdot \cos 270^{\circ})$

$D'(x,y) = (8, 0)$

$E'(x,y) = (6\cdot \cos 270^{\circ}-5\cdot \sin 270^{\circ}, 6\cdot \sin 270^{\circ}+5\cdot \cos 270^{\circ})$

$E'(x,y) = (5, -6)$

$F'(x,y) = (5\cdot \cos 270^{\circ}-3\cdot \sin 270^{\circ}, 5\cdot \sin 270^{\circ}+3\cdot \cos 270^{\circ})$

$F'(x,y) = (3, -5)$

Nota - The statement is incomplete. Complete form is presented below:

Given the rectangle CDEF with vertices C(-1, 6), D(0, 8), E(6, 5) and F(5, 3), find the rectangle CDEF by rotating 270° counterclockwise.

We kindly invite to check this question on rotation: https://brainly.com/question/1571997