A counterclockwise quarter-turn Q about the origin is applied to the point (x, y). What are the coordinates of the image point? Answer in the form Q(x, y) = (ax + by, cx + dy).
Well, let's see. If we apply a quarter-turn counterclockwise about the origin, it means we're rotating the point (x, y) by 90 degrees. So, the image point (Q(x, y)) would have coordinates (-y, x).
But wait, why use all those fancy letters and symbols? Let me simplify it for you:
Q(x, y) = (-y, x)
So, after the counterclockwise quarter-turn about the origin, the coordinates of the image point are (-y, x). Voila!
To find the coordinates of the image point after a counterclockwise quarter-turn Q about the origin is applied to the point (x, y), we need to use the transformation matrix for a counterclockwise rotation of 90 degrees.
The transformation matrix for a counterclockwise rotation of 90 degrees about the origin is:
| 0 -1 |
| 1 0 |
Let's denote this matrix as M:
M = | 0 -1 |
| 1 0 |
To find the image point (x', y'), we multiply the vector (x, y) by the transformation matrix M:
| x' | | 0 -1 | | x |
| y' | = | 1 0 | * | y |
Expanding the matrix multiplication, we get:
x' = 0*x + (-1)*y = -y
y' = 1*x + 0*y = x
Therefore, the coordinates of the image point (x', y') after a counterclockwise quarter-turn Q about the origin are given by:
Q(x, y) = (-y, x)
To determine the coordinates of the image point after a counterclockwise quarter-turn Q about the origin, we can use a rotation matrix. The rotation matrix for a counterclockwise quarter-turn, which is 90 degrees, is:
[0 -1]
[1 0]
To find the coordinates of the image point, we multiply this rotation matrix by the original coordinates of the point (x, y) as follows:
[x' y'] = [0 -1] * [x]
[y]
= [(0 * x) + (-1 * y)]
[(1 * x) + (0 * y)]
= [-y]
[x]
Thus, the coordinates of the image point (x', y') after the counterclockwise quarter-turn Q about the origin are given by:
Q(x, y) = (-y, x)
If by a quarter-turn you mean a counterclockwise roation of 90°, then that would simply be
(x,y) ----> (-y,x)
You can illustrate that by drawing a right-angled triangle in the first quadrant, with sides +x and +y
Now do a 90° rotation to put the triangle in the second quadrant.
Notice that the base is now y, but of course it is going to the left, so -y
and the height is the original x of the first triangle.