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Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.
At the circus, Mike the magnificent is walking the tight rope. It takes him 10 equal size steps to get across the rope. He takes seven steps flawlessly, then wobbles a bit, and quickly takes the last three steps to land safely on the end platform.
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It just so happens that Mike just performed a mathematical feat called partitioning a line segment. Partitioning a directed line segment, AB, into a ratio a/b involves dividing the line segment into a + b equal parts and finding a point that is a equal parts from A and b equal parts from B.
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Consider Mike again. The point at which he wobbled on the tight rope (a line segment of 10 equal parts) is seven equal parts from the start and three equal parts from the end, so the point at which Mike wobbled partitioned the line segment into the ratio 7/3.
Partitioning a directed line segment seems simple enough, but what if we are given the two endpoints of a directed line segment, and want to find the point that partitions the line segment into the ratio a/b?
Thankfully, we can do this fairly easily using parts of the slope of the line segment. The slope of the line segment with endpoints (x1, y1) and (x2, y2) gives us the rate at which y is changing with respect to x, and we can find it using the slope formula:
Slope = Rise/Run = (Change in y)/(Change in x) = (y2 - y1)/(x2 - x1)
If we are given a line segment AB, where A = (x1, y1) and B = (x2, y2), and we want to partition it into the ratio a/b, then we want to find a point P that falls a equal parts from point A and b equal parts from point B on the line segment. We can do this using the following steps:
These steps also give way to a nice easy formula for P:
P = (x1 + c(x2 - x1), y1 + c(y2 - y1))
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Hmmm…that seems to make sense, but don't you think an example will make things even more clear?
Suppose we have a directed line segment AB, where A = (1,2) and B = (8,7), and we want to partition it with the ratio 3/5. In other words, we want to find a point, P, that is three equal parts from A and is five equal parts from B. Let's take it through our steps, and then we'll verify our answer with our formula.
The first thing we want to do is find the ratio c that tells us what fraction of the way P is from A to B. To do this, we use our formula c = a/ (a + b), where a = 3 and b = 5.
Okay, so we know P will be 3/8 of the way from A to B. Now, we want to find the rise and the run of the slope of the line segment. Again, we can use our formulas with the points (1,2) and (8,7).
Alright, one more step! We need to add câ‹…(run) to x1, and add câ‹…(rise) to y1. This adds 3/8 of the change in x from A to B to the x-coordinate of A and 3/8 of the change in y from A to B to the y-coordinate of A, giving our desired point P that is 3/8 of the way from A to B and partitions the line into the ratio 3/5.
This gives that the point P is (3.625, 3.875). Let's see if our formula gives the same thing:
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It does! Let's take a look at the point P on AB on a graph.
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Sure enough, the point P = (3.625, 3.875) partitions AB into the ratio 3/5.
Partitioning a line segment, AB, into a ratio a/b involves dividing the line segment into a + b equal parts and finding a point that is a equal parts from A and b equal parts from B.
When finding a point, P, to partition a line segment, AB, into the ratio a/b, we first find a ratio c = a / (a + b).
The slope of a line segment with endpoints (x1, y1) and (x2, y2) is given by the formula rise/run, where:
Given a line segment AB and a partitioning ratio a/b, we can find the point P that partitions the line segment appropriately using a formula that involves parts of the slope of the line segment. That is:
This formula makes partitioning a line segment a cinch!
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Explorations in Core Math - Geometry: Online Textbook Help13 chapters | 127 lessons
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