Given point A (2, 4) on the triangle Quadrant I, where is the new point if the triangle is rotated 270 degrees counterclockwise?
AAAaannndd the bot gets it wrong yet again!
(4,-2)
What on earth is that bot's problem?
Most of the time it is wrong.
To find the new location of point A after rotating the triangle 270 degrees counterclockwise, you can use a transformation matrix.
First, let's plot the original point A (2, 4) in Quadrant I:
```
Y
|
|
|
A(2,4) |________ X
```
After a 270 degrees counterclockwise rotation, the point A will be located in Quadrant IV. To visualize this, let's create a new coordinate axis after the rotation:
```
A'(?,?) |________ X
|
|
|
Y
```
To find the coordinates of A', we need to apply the transformation matrix for a 270 degrees counterclockwise rotation about the origin:
```
| x' | | cos(270°) -sin(270°) | | x |
| y' | = | sin(270°) cos(270°) | * | y |
```
Since cos(270°) = 0 and sin(270°) = -1, we can simplify the transformation matrix:
```
| x' | | 0 -1 | | x |
| y' | = | 1 0 | * | y |
```
Applying the matrix to the original point A (2, 4):
```
| x' | | 0 -1 | | 2 |
| y' | = | 1 0 | * | 4 |
```
Performing the matrix multiplication:
```
| x' | | 0 -1 | | 2 | | 0*2 + (-1)*4 | | -4 |
| y' | = | 1 0 | * | 4 | = | 1*2 + 0*4 | = | 2 |
```
Therefore, the new location of point A after rotating the triangle 270 degrees counterclockwise is A'(-4, 2).