Geometry and Trigonometry - Rotation around a given point

Rotate the point $A(3, 5)$ by $90^{\circ}$ counter-clockwise about the origin $(0, 0)$.

$A(3, 5) \rightarrow A'(-5, 3)$

Using the rule for $90^{\circ}$ counter-clockwise rotation $(x, y) \rightarrow (-y, x)$, we take the $y$-coordinate $5$, negate it to get $-5$ (the new $x$), and take the original $x$-coordinate $3$ to be the new $y$.

Rotate the point $P(4, 2)$ by $180^{\circ}$ about the point $C(1, 1)$.

$P(4, 2) \xrightarrow{\text{Translate}} (4-1, 2-1) = (3, 1)$ $(3, 1) \xrightarrow{\text{Rotate } 180^{\circ}} (-3, -1)$ $(-3, -1) \xrightarrow{\text{Translate back}} (-3+1, -1+1) = (-2, 0)$ $P' = (-2, 0)$

First, we translate the center of rotation to the origin by subtracting $C(1, 1)$ from $P(4, 2)$, resulting in $(3, 1)$. Next, we apply the $180^{\circ}$ rotation rule $(x, y) \rightarrow (-x, -y)$ to get $(-3, -1)$. Finally, we translate back by adding $C(1, 1)$ to get the final coordinates $(-2, 0)$.

1. Apply the rule for $90^{\circ}$ clockwise rotation: $(x, y) \rightarrow (y, -x)$.
2. Calculate the new coordinates:
   • $A(1, 1) \rightarrow A'(1, -1)$
   • $B(4, 1) \rightarrow B'(1, -4)$
   • $C(1, 3) \rightarrow C'(3, -1)$
3. The vertices of the rotated triangle are $A'(1, -1)$, $B'(1, -4)$, and $C'(3, -1)$.

A clockwise rotation of $90^{\circ}$ moves points from the first quadrant to the fourth quadrant. The $x$ and $y$ values are swapped, and the new $y$-coordinate becomes negative.

1. Translate the center $M(2, 2)$ to the origin by subtracting its coordinates: $L_{shifted} = (5-2, 4-2) = (3, 2)$.
2. Apply the $90^{\circ}$ CCW rotation rule $(x, y) \rightarrow (-y, x)$ to the shifted point: $(3, 2) \rightarrow (-2, 3)$.
3. Translate back by adding the coordinates of $M$: $L' = (-2+2, 3+2) = (0, 5)$.
4. The final coordinates are $L'(0, 5)$.

When the center is not the origin, we temporarily treat the center as $(0,0)$, rotate, and then shift back to the original position.