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Parametrizing a Line Segment - Problem 2
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Finding parametric equations of line segments. Here’s an example; Let point A be -9, 4 and point B be 7, -8. Find parametric equations for segment AB, going from B to A. So this raises the issue that when you are parametrizing a line segment, you can parametrize it so that as t advances you go from one point to the other or vice versa.

So we are going to start form point B. And if B is our starting point, then we need to look at these numbers 7 -8. So remember that our formula is x equals 1 minus t, times the starting x value and that’s 7. Plus t times the finishing x value and that’s -9. And y is 1 minus t times the starting y value, that’s -8, plus t times the ending y value and that’s 4.

So these would be our equations for line segment. Just remember that you have to have the domain restriction; t is between 0 and 1. So when t equals 0, you are going to get the first point 7 -8. When t equals 1, you are going to get the final point -9, 4. Then you get all the points in between when t is between 0 and 1.

Now let’s find parametric equations for AB going from A to B. So it’s going to be very similar, it’s just you are going to switch A and B. So this is going to be our starting point. So this is our starting x value -9, it goes with 1 minus t. And then plus t times, and our finishing x value is 7. Then for y, we have 1 minus t times our starting y value of 4, plus t times our finishing y value of -8. And again, you still need to have the domain restriction t is between 0 and 1.

So you just get the line segment. Otherwise you get a line going through points A and B. Part c is rather interesting. It says find the point Q which is 3/4 the way from A to B. We should use this parametrization of segment AB, because this one goes from A to B. And so if I plug in 3/4 I’ll actually get the point that it's 3/4 the way from A to B.

So that’s what I’m going to do. So I plug in 3/4 here, and I get 1 minus 3/4 which is ¼ times -9. Plus t is 3/4 times 7. And the y value ¼ times 4, plus 3/4 times -8.

So I just have to multiply these out. ¼ times -9, -9/4. Plus 21/4, so 21 minus 9 is 12, 12/4 is 3. And ¼ of 4 is 1, and 3/4 of -8 is -6. So I get -5. And so point Q that I’m looking for, has coordinates 3,-5. And that’s it.

If I had been looking for a point Q which is 3/4 away from B to A, I would have used this parametrization, because this one goes from B to A. But that’s how you find a point that’s in between points A and B.

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