Given point A (2, 4) on the triangle in quadrant 1 how would you describe the coordinates of the new point when the triangle is rotated 90 degrees clockwise
To describe the coordinates of the new point after rotating the triangle 90 degrees clockwise, we can use the rotation matrix:
\[
\begin{bmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta)
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
\]
Since we are rotating the triangle 90 degrees clockwise, the angle of rotation, θ, is -90 degrees or -π/2 in radians. Applying the rotation matrix to the coordinates of point A (2, 4):
\[
\begin{bmatrix}
\cos(-\frac{\pi}{2}) & -\sin(-\frac{\pi}{2}) \\
\sin(-\frac{\pi}{2}) & \cos(-\frac{\pi}{2})
\end{bmatrix}
\begin{bmatrix}
2 \\
4
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
2 \\
4
\end{bmatrix}
\]
Simplifying:
\[
\begin{bmatrix}
0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
2 \\
4
\end{bmatrix}
=
\begin{bmatrix}
4 \\
-2
\end{bmatrix}
\]
Therefore, the new coordinates of point A after rotating the triangle 90 degrees clockwise are (4, -2).