The formula for rotating 90 degrees counterclockwise about the origin is ((x,y)\rightarrow(-y,x)). Put another way, the original (y)-coordinate becomes the negative (x)-coordinate, while the original (x)-coordinate becomes the new (y)-coordinate That's the whole idea..
Introduction
A rotation turns a point, line, or shape around a fixed point called the center of rotation. When the rotation is (90^\circ) counterclockwise, every point moves one quarter of a full turn in the positive angular direction.
In the standard Cartesian coordinate plane, where the (x)-axis points right and the (y)-axis points upward, the rule is:
[ \boxed{(x,y)\rightarrow(-y,x)} ]
This formula assumes that the center of rotation is the origin, ((0,0)) That's the part that actually makes a difference..
For example:
[ (4,2)\rightarrow(-2,4) ]
The original coordinates are swapped, and the new (x)-coordinate is negative.
The Basic Formula
To rotate any point (90^\circ) counterclockwise around the origin:
- Start with the point ((x,y)).
- Swap the coordinates to obtain ((y,x)).
- Change the sign of the new (x)-coordinate.
- The result is ((-y,x)).
Examples
| Original Point | Rotated Point |
|---|---|
| ((1,0)) | ((0,1)) |
| ((0,1)) | ((-1,0)) |
| ((3,5)) | ((-5,3)) |
| ((-2,4)) | ((-4,-2)) |
| ((-6,-1)) | ((1,-6)) |
A helpful pattern is:
[ \boxed{\text{First coordinate}=-\text{old }y} ]
[ \boxed{\text{Second coordinate}=\text{old }x} ]
Why the Formula Works
A (90^\circ) counterclockwise rotation moves each point to a position perpendicular to its original position. The distance from the origin remains unchanged.
Consider the point ((3,0)), which lies three units to the right of the origin. After a (90^\circ) counterclockwise rotation, it should lie three units above the origin:
[ (3,0)\rightarrow(0,3) ]
Now consider ((0,3)), which lies three units above the origin. Rotating it another (90^\circ) counterclockwise moves it three units to the left:
[ (0,3)\rightarrow(-3,0) ]
The formula ((x,y)\rightarrow(-y,x)) produces both results correctly.
Rotation Using a Matrix
A (90^\circ) counterclockwise rotation can also be represented with a rotation matrix:
[ R_{90^\circ}= \begin{bmatrix} 0 & -1\ 1 & 0 \end{bmatrix} ]
Multiplying this matrix by a coordinate column vector gives:
[ \begin{bmatrix} 0 & -1\ 1 & 0 \end{bmatrix} \begin{bmatrix} x\ y \end{bmatrix}
\begin{bmatrix} -y\ x \end{bmatrix} ]
Therefore:
[ \begin{bmatrix} x\ y \end{bmatrix} \rightarrow \begin{bmatrix} -y\ x \end{bmatrix} ]
This matrix form is especially useful in geometry, physics, engineering, robotics, and computer graphics Still holds up..
Rotating a Shape
To rotate an entire shape, rotate each of its vertices using the same formula. After transforming all vertices, connect them in the same order as the original shape Worth keeping that in mind..
Example
Example
Consider triangle (ABC) with vertices (A(1,1)), (B(3,1)), and (C(2,3)). To rotate this triangle (90^\circ) counterclockwise about the origin, apply the formula ((x,y)\rightarrow(-y,x)) to each vertex:
[ \begin{aligned} A(1,1) &\rightarrow A'(-1,1)\[4pt] B(3,1) &\rightarrow B'(-1,3)\[4pt] C(2,3) &\rightarrow C'(-3,2) \end{aligned} ]
Plotting (A'), (B'), and (C') and connecting them in the same order produces the rotated triangle. The shape, side lengths, and angles are all preserved — only the orientation has changed.
| Vertex | Original | Rotated |
|---|---|---|
| (A) | ((1,1)) | ((-1,1)) |
| (B) | ((3,1)) | ((-1,3)) |
| (C) | ((2,3)) | ((-3,2)) |
Notice that the triangle has effectively turned one quarter-turn in the positive angular direction. The counterclockwise orientation of the vertices is maintained, confirming that the rotation is a rigid transformation — it preserves distances, angles, and the overall structure of the figure That's the whole idea..
Rotating Around a Point Other Than the Origin
The formula ((x,y)\rightarrow(-y,x)) applies directly only when the center of rotation is the origin. When the center of rotation is an arbitrary point ((h,k)), a three-step process is used:
- Translate the figure so that the center of rotation moves to the origin. Subtract ((h,k)) from every point: ((x,y)\rightarrow(x-h,;y-k)).
- Apply the standard (90^\circ) counterclockwise rotation: ((x-h,;y-k)\rightarrow(-(y-k),;x-h)).
- Translate back by adding ((h,k)) to every rotated point: ((-(y-k)+h,;x-h+k)).
The combined rule becomes:
[ (x,y);\rightarrow;(h-k+y+k,;x-h+k-k) ]
which simplifies to:
[ \boxed{(x,y);\rightarrow;(h-y+k,;x-h+k)} ]
Wait — let me correct that carefully. After step 2 we have the point ((-(y-k),;x-h)). Adding ((h,k)) back gives:
[ \big(-(y-k)+h,;;x-h+k\big) = (h-y+k,;;x-h+k) ]
Hmm, let me recheck. Which means (-(y-k) = -y+k), so the first coordinate is (-y+k+h = h-k+y). Wait, that's (h + k - y) Simple as that..
First coordinate: (-(y - k) + h = -y + k + h = h + k - y)
Second coordinate: ((x - h) + k = x - h + k)
So the correct combined rule is:
[ \boxed{(x,y);\rightarrow;(h+k-y,;;x-h+k)} ]
Quick verification: Rotate the point ((5,3)) by (90^\circ) counterclockwise around the center ((2,1)).
- Translate: ((5-2,;3-1) = (3,2))
- Rotate: ((-2,3
