# Rotation

## Determining new coordinates using a picture

You can determine the new coordinates of an object

Only if

the origin is (0,0)

and only if

you rotate `90^@`, `180^@` or `270^@`

Remember this picture:

The above says:

- Move clockwise 90°
- Swap the coordinates
- Then multiply the second coordinate by `-1`

- Move anticlockwise 90°
- Swap the coordinates
- Then multiply the first coordinate by `-1`

- Move 180°
- Just change both signs

**Example 1**

With origin (0,0) rotate the point (2,3) by

- Clockwise 90°
- Anticlockwise 90°
- Through 180°

and plot your answers

**Answer:**

To remember how to do this write out the following:

- Clockwise 90°
- Swap the coordinates:

The coordinates are`(2,3)`

swap and they become `(3,2)` - Then multiply the second coordinate by `-1`:

`(3,2)` becomes `(3,-2)`

- Swap the coordinates:
- Anticlockwise 90°
- Swap the coordinates:

The coordinates are`(2,3)`

swap and they become `(3,2)` - Then multiply the first coordinate by `-1`:

`(3,2)` becomes `(-3,2)`

- Swap the coordinates:
- Through 180°
- Just change the signs:

The coordinates are `(2,3)`

`(2,3)` becomes `(-2,-3)`

- Just change the signs:

Plot the answer:

**Example 2**

Rotate The object below anticlockwise by 90° with origin (0,0).

So we would use: SWAP and 1st number `times-1`

Point A `=(-5,-1)` becomes `(-1(times-1),-5)=(1,-5)`

Point B `=(-2,-1)` becomes `(-1(times-1),-2)=(1,-2)`

Point C `=(-4,-3)` becomes `(-3(times-1),-4)=(3,-4)`

The graph becomes:

Does this look correct? YES IT DOES

**Example 3**

Rotate the triangle below anticlockwise by 90° with origin (0,0).

Use

We would use: SWAP and 1st number `times-1`

`A=(1,2)` becomes `(2times(-1),1)=(-2,1)`

`B=(3,1)` becomes `(1times(-1),3)=(-1,3)`

`C=(2,-2)` becomes `(-2times(-1),2)=(2,2)`

The graph becomes:

**Example 4**

Rotate the line below by 180° with the origin at (0,0).

Use

We would use: Change both signs

So if `A=(-3,2)` and `B=(2,3)`

The new coordinates would be:

`A^1=(3,-2)` and `B^1=(-2,-3)`

Plot the results:

As a check you could run `A` and `B` through the origin and see if they hit`A^1` and `B^1` at the same distance from the origin:

This is correct.