If we enlarge (or reduce) an object only in one direction, it is called a stretch!
Mathematically speaking, there are two parameters we need to know:
- Stretch Factor: how much we stretch by. Usually given the value k
- Invariant Line: the line which we stretch away from. Think of it as the ‘unchanging line’.
Example:
If the invariant line is the y-axis, only the x-coordinate is affected by the stretch factor k.
(x,y) –> (kx,y)
And vice cersa for if the invariant line is the x-axis, only the y-coordinate changes.
(x,y) –> (x,ky)
Stretching with the x or y-axes as the IL is easy enough. However, if the IL is slightly off the x or y-axes, you have to count the distance of each point from the IL. Then you stretch according to that distance.
Look at the example below:
- P has the coordinate (4,0). Count: how far from the IL is P?
Answer…P is 2 units away from the IL! - What is the stretch factor? k=1/2
- That means I multiply the distance 2 by 1/2. Obviously, that will be 1.
- Therefore, P stretched with a factor of k=1/2 would have a new position of (4, -1).
Why -1 ? Because the stretch factor is less than 1, so the stretch is going to be a stretching towards the IL. Like it’s squishing downwards!
P(4,0) –> P'(4,-1)
Let’s try point Q, then you should be able to do points R and S yourself.
As you can see, point Q is below the IL. Doesn’t matter. Just count the distance away again.
- The coordinate of Q is (8,-5).
- That means Q is 3 units away from the IL. The stretch factor is 1/2, so the new distance has to be 1.5 units away from the IL.
- Only the y-coordinate will be stretched by k=1/2. So the new position will be (8,-3.5)
Q(8,-5) –> Q'(8,-3.5)
Noice.
Try R and S yourself, and check the answer below!
Now, the real challenge is in trying to solve these kinds of problems without drawings. You can try to visualise it, or draw it based on the info given. For example:
Answers…coming soon!
Hype Maths 