Question

trapezoid ABCD is shown on the graph if the trapezoid is rotated 90 degrees counterclockwise about the origin and the resulting image is translated 6 unit left, what are the coordinates of C

Answer 1

Answer:

3

Step-by-step explanation:

Answer 2

The coordinates of C'' is $C''(x,y) = (-7,4)$.

In this question we must rotate every point of the trapezoid ABCD around the origin, which is defined by the following vectorial formula:

$P'(x,y) = O(x,y) + (r_{OP, x}\cdot \cos \theta - r_{OP,y}\cdot \sin \theta, r_{OP,x}\cdot \sin \theta +r_{OP,y}\cdot \cos \theta)$ (1)

Where:

• $O(x,y)$ - Center of rotation.
• $r_{OP}$ - Norm of the vector OP.
• $\theta$ - Angle of rotation, in sexagesimal degrees.

Later, we apply the following translation:

$P''(x,y) = P'(x,y) + (-6,0)$ (2)

If we know that $O(x,y) = (0,0)$, $r_{OC,x} = 4$, $r_{OC,y} = 1$, $\theta = 90^{\circ}$ and $C(x,y) = (4,1)$, the coordinates of C'' are:

$C'(x,y) = (0,0) + (4\cdot \cos 90^{\circ}-1\cdot \sin 90^{\circ}, 4\cdot \sin 90^{\circ}+1\cdot \cos 90^{\circ})$

$C'(x,y) = (-1, 4)$

$C''(x,y) = (-1,4) +(-6,0)$

$C''(x,y) = (-7,4)$

The coordinates of C'' is $C''(x,y) = (-7,4)$.

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