Given a point (1, 2) on a geometric figure what is the new point when the figure is rotated counterclockwise about the origin 90 degrees?
To rotate a point counterclockwise about the origin 90 degrees, we can use the following rotation formula:
(x', y') = (x * cos θ - y * sin θ, x * sin θ + y * cos θ)
where (x, y) represents the original point and (x', y') represents the new point after rotation.
For a 90-degree counterclockwise rotation, we have θ = 90 degrees or θ = π/2 radians.
Applying the formula to the point (1, 2):
(x', y') = (1 * cos(π/2) - 2 * sin(π/2), 1 * sin(π/2) + 2 * cos(π/2))
= (1 * 0 - 2 * 1, 1 * 1 + 2 * 0)
= (-2, 1)
Therefore, the new point after a 90-degree counterclockwise rotation about the origin is (-2, 1).
To find the new point after rotating a point counterclockwise 90 degrees about the origin, we can use the following rotation matrix:
\[ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \]
In this case, the angle of rotation \(\theta\) is 90 degrees. Let's substitute the values of the point (1, 2) into the rotation matrix:
\[ \begin{bmatrix} \cos(90^\circ) & -\sin(90^\circ) \\ \sin(90^\circ) & \cos(90^\circ) \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
Next, let's evaluate the trigonometric functions for a 90-degree angle:
\[ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} \]
Now, multiply the matrices:
\[ \begin{bmatrix} 0(1) + (-1)(2) \\ 1(1) + 0(2) \end{bmatrix} \]
Simplifying:
\[ \begin{bmatrix} -2 \\ 1 \end{bmatrix} \]
Therefore, when the figure is rotated counterclockwise about the origin 90 degrees, the new point for (1, 2) would be (-2, 1).