45-Degree Rotations About the Origin
Okay, so we're going to pick up the topic of transformations where we left it in this post.
You'll all remember from that post (nearly a year ago) that we "derived" from a few examples a simple rule we could use to determine the coordinates of the image of a point rotated 90 degrees (counterclockwise) about the origin: the x-coordinate becomes the new y-coordinate, and the negative y-coordinate becomes the new x-coordinate.
As a general rule, if we take our starting coordinates to be (x, y), then there are really only two sets of coordinates to remember for degree rotations about the origin that are multiples of 90 . . .
For clockwise rotations, obviously we just transpose these sets of coordinates.
Take a look at the line graphed below (in red). The measures of the angles formed in the first and third quadrants by this line and the x-axis of the coordinate grid are each 45o.
It is important for you to see that for each point on this red line, the x- and y-coordinates are equal. That is, the coordinates for each point on the red line can be represented as (x, x) or (y, y). This is fairly obvious, since we can see that the line passes through the points (1, 1), (2, 2), etc. But keep in mind that (0.0015, 0.0015), (4,501.2, 4,501.2), etc., are also points on the line.
Something else that is extremely important to see is that, because of the all-powerful Pythagorean equation, the diagonal (d) formed by the red line inside each grid square has a length equal to the square root of 2, which is approximately 1.414.
Okay, so that was the easy part.
As I mentioned, we will approach 45o rotations in the same way that we approached 90o rotations. So let's rotate the point at (3, 2) 45o about the origin.
We take our "claw," which is connected to the point located at (3, 2) and rotate it 45o. Then we take a look at each "arm" of the claw.
Right away, we find the situation to be a bit trickier than the one we encountered with 90o rotations. Where are the endpoints of the rotated arms? We can see that the end of the rotated x-arm is pretty close to (2, 2), but not exactly there, and it is even harder to guess where exactly the end of the rotated y-arm is. So, just like that, we need to once again pause and enjoy a tall, frosty mug of fresh perspective.
Take a look again at the red line and consider what we know:
(1) no matter where on the line we draw a point, the x- and y-coordinates of that point will be equal, and (2) the diagonal formed inside each grid square has a length equal to the square root of 2. What (1) and (2) indicate is that there is a functional relationship between the length of any diagonal line segment and the horizontal or vertical location of one of its endpoints on the coordinate grid.
How might we describe this relationship? Here's how (at least for the first quadrant):
So, let's look again at our claw, attached to the point at (3, 2) and rotated 45o about the origin. We know the length of the diagonal line segment formed by this rotation. It's simply the x-arm of our claw, which has a length of 3.
Therefore, the location (both the x- and y-coordinates) of the end of the x-arm can be found by dividing the length of the arm, 3, by the square root of two:
To make the next section a little clearer, we can write the coordinates of the end of the rotated x-arm this way:
When we write the coordinates in this way, we can more easily see them as vectors--that is, they are numbers that describe not only magnitude, but direction as well. On the coordinate grid, movement to the right on the x-axis can be represented by a positive number, and movement to the left can be represented by a negative number. Similarly, on the y-axis, movement up can be represented by a positive number, and movement down can be represented by a negative number. If we begin at the origin (0, 0), then to find the location of the end of the rotated x-arm, we follow the x-axis to the right (positive) and follow the y-axis up (positive). This is why both of the coordinates that describe the location of the end of the x-arm are positive numbers.
Anyway, obviously the location of the end of the x-arm is not what we want. We want to know the location of the end of the y-arm of our "claw." This is the location of the point (3, 2) rotated 45 degrees about the origin.
As it turns out, all the hard work is behind us now, because the line that contains the rotated y-arm segment also forms a 45-degree angle with the x-axis. (There are probably dozens, if not hundreds of ways to show this, and I'll leave it up to curious readers to tackle it themselves.)
Since the rotated y-arm is also oriented at a 45-degree angle, we can treat it the same way we treated the x-arm. To make this as clear as possible, let's first detach the rotated y-arm from the rotated x-arm and make its "starting point" (0, 0):
If we take the length of this arm, 2, and divide by the square root of 2, we will come up with this location for the end of the detached y-arm:
But, of course, this isn't right, because we did not take into account direction. We still follow the y-axis up to find the end of this arm, so the y-value should be positive but, in this case, we follow the x-axis to the left to find the end of the arm. This means that the x-value should be negative:
So, take a look again at the rotation. We see that the x-arm moves up and to the right, while the y-arm moves up and to the left.
To find the location of the end of the y-arm, we use the end of the x-arm as a starting point and simply add the coordinates:
All that's left is to notice that in the numerators of the fractions above, we see the coordinates of the original point (3, 2). Play around with the grid below to see that the rule is true. The coordinates of a point (x, y) rotated 45 degrees (counterclockwise) about the origin are:
What is not immediately obvious is why this rule should hold no matter what quadrants are involved. For example, the 45-degree rotation in the second quadrant has a down y-movement, which is not reflected in the rule. It's pretty simple to figure out how this works after you play with the rotations a bit (or, of course, you can figure it out algebraically too). Enjoy.
You'll all remember from that post (nearly a year ago) that we "derived" from a few examples a simple rule we could use to determine the coordinates of the image of a point rotated 90 degrees (counterclockwise) about the origin: the x-coordinate becomes the new y-coordinate, and the negative y-coordinate becomes the new x-coordinate.
As a general rule, if we take our starting coordinates to be (x, y), then there are really only two sets of coordinates to remember for degree rotations about the origin that are multiples of 90 . . .
90-degree (counterclockwise) rotation: (-y, x). . . assuming one needs no help remembering that the coordinates of a point under a 180-degree rotation simply take opposite signs (-x, -y).
270-degree (counterclockwise) rotation: (y, -x)
For clockwise rotations, obviously we just transpose these sets of coordinates.
90-degree (clockwise) rotation: (y, -x)
270-degree (clockwise) rotation: (-y, x)
45-Degree Rotations
We can use the same kind of thinking to determine a general rule for 45-degree rotations as well. But first--and again--we need to get the right perspective (or, rather, a useful perspective).Take a look at the line graphed below (in red). The measures of the angles formed in the first and third quadrants by this line and the x-axis of the coordinate grid are each 45o.
It is important for you to see that for each point on this red line, the x- and y-coordinates are equal. That is, the coordinates for each point on the red line can be represented as (x, x) or (y, y). This is fairly obvious, since we can see that the line passes through the points (1, 1), (2, 2), etc. But keep in mind that (0.0015, 0.0015), (4,501.2, 4,501.2), etc., are also points on the line.
Something else that is extremely important to see is that, because of the all-powerful Pythagorean equation, the diagonal (d) formed by the red line inside each grid square has a length equal to the square root of 2, which is approximately 1.414.
Okay, so that was the easy part.
As I mentioned, we will approach 45o rotations in the same way that we approached 90o rotations. So let's rotate the point at (3, 2) 45o about the origin.
We take our "claw," which is connected to the point located at (3, 2) and rotate it 45o. Then we take a look at each "arm" of the claw.
Right away, we find the situation to be a bit trickier than the one we encountered with 90o rotations. Where are the endpoints of the rotated arms? We can see that the end of the rotated x-arm is pretty close to (2, 2), but not exactly there, and it is even harder to guess where exactly the end of the rotated y-arm is. So, just like that, we need to once again pause and enjoy a tall, frosty mug of fresh perspective.
Take a look again at the red line and consider what we know:
(1) no matter where on the line we draw a point, the x- and y-coordinates of that point will be equal, and (2) the diagonal formed inside each grid square has a length equal to the square root of 2. What (1) and (2) indicate is that there is a functional relationship between the length of any diagonal line segment and the horizontal or vertical location of one of its endpoints on the coordinate grid.
How might we describe this relationship? Here's how (at least for the first quadrant):
Once we know the length of any (45o) diagonal line segment, in order to find the horizontal or vertical location of a specific endpoint, we can simply divide the length by the square root of 2.
So, let's look again at our claw, attached to the point at (3, 2) and rotated 45o about the origin. We know the length of the diagonal line segment formed by this rotation. It's simply the x-arm of our claw, which has a length of 3.
Therefore, the location (both the x- and y-coordinates) of the end of the x-arm can be found by dividing the length of the arm, 3, by the square root of two:
To make the next section a little clearer, we can write the coordinates of the end of the rotated x-arm this way:
When we write the coordinates in this way, we can more easily see them as vectors--that is, they are numbers that describe not only magnitude, but direction as well. On the coordinate grid, movement to the right on the x-axis can be represented by a positive number, and movement to the left can be represented by a negative number. Similarly, on the y-axis, movement up can be represented by a positive number, and movement down can be represented by a negative number. If we begin at the origin (0, 0), then to find the location of the end of the rotated x-arm, we follow the x-axis to the right (positive) and follow the y-axis up (positive). This is why both of the coordinates that describe the location of the end of the x-arm are positive numbers.
Anyway, obviously the location of the end of the x-arm is not what we want. We want to know the location of the end of the y-arm of our "claw." This is the location of the point (3, 2) rotated 45 degrees about the origin.
As it turns out, all the hard work is behind us now, because the line that contains the rotated y-arm segment also forms a 45-degree angle with the x-axis. (There are probably dozens, if not hundreds of ways to show this, and I'll leave it up to curious readers to tackle it themselves.)
Since the rotated y-arm is also oriented at a 45-degree angle, we can treat it the same way we treated the x-arm. To make this as clear as possible, let's first detach the rotated y-arm from the rotated x-arm and make its "starting point" (0, 0):
If we take the length of this arm, 2, and divide by the square root of 2, we will come up with this location for the end of the detached y-arm:
But, of course, this isn't right, because we did not take into account direction. We still follow the y-axis up to find the end of this arm, so the y-value should be positive but, in this case, we follow the x-axis to the left to find the end of the arm. This means that the x-value should be negative:
So, take a look again at the rotation. We see that the x-arm moves up and to the right, while the y-arm moves up and to the left.
To find the location of the end of the y-arm, we use the end of the x-arm as a starting point and simply add the coordinates:
All that's left is to notice that in the numerators of the fractions above, we see the coordinates of the original point (3, 2). Play around with the grid below to see that the rule is true. The coordinates of a point (x, y) rotated 45 degrees (counterclockwise) about the origin are:
What is not immediately obvious is why this rule should hold no matter what quadrants are involved. For example, the 45-degree rotation in the second quadrant has a down y-movement, which is not reflected in the rule. It's pretty simple to figure out how this works after you play with the rotations a bit (or, of course, you can figure it out algebraically too). Enjoy.
Labels: mathematics