Text description - screencast for exercise 1

[The narrator reads out the onscreen text.]

NARRATOR: Exercise 1. If a point, P, with coordinates (x, y) lies on the line AB where A has coordinates (x1, y1) and B has coordinates (x2, y2) and the ratio of AP to PB is a to b with a and b both positive, show that x, that is the x-coordinate of the point P, is equal to b times x1 plus a times x2 all divided by a plus b, and y, that is the y-coordinate of the point P, is equal to b times y1 plus a times y2 all divided by a plus b.

[A diagram is labelled A P to P B equals a to b. In the diagram a diagonal line stretches between the point A (x1, y1) and the point B (x2, y2). The point P (x, y) is marked about two-thirds of the distance between A and B. The line AB forms the hypotenuse of a right angled triangle. The other two sides are dotted lines. The vertical side is marked y2 minus y1, the horizontal side is marked x2 minus x1.]

NARRATOR: Let's first consider the situation geometrically. And so we have the line AB with the point P on that line. And the point P divides the line in the ratio A to B.

[The line between A and P is marked with a lower-case a. The line between P and B is marked with a lower-case b.]

NARRATOR: So for example, we could say that the distance from A to P is a and the distance from P to B is b. And so that means the total distance from A to B is a plus b.

[The third corner of the triangle is marked B-dash. A line runs down from P to the line A B-dash. The intersection is marked P-dash (x2 minus x1). A second right angled triangle is marked between A, P and P-dash. A second equation reads A P-dash to P-dash B-dash equals a to b]

NARRATOR: Now if we consider the two triangles that we have here, we note first that they are in fact similar triangles, so A P P-dash is similar to A B B-dash due to the fact that the two triangles have the same angles. And we also note that we now have a triangle with a hypotenuse of A that's similar to another triangle with a hypotenuse a plus b. And so we can work out some scale factors here. So if we were to look at the scale factor going from the smaller triangle to the larger triangle, we would be multiplying the side lengths by a plus b all divided by a. And if we were to go from the larger triangle to the smaller one, we would be multiplying the side lengths by a divided by a plus b. And so using that idea, then, we see that if we look at the horizontal distances, so dealing with the x-coordinates, we see that A to P-dash would be equal to a divided by a plus b multiplied by A to B-dash.

And we can use this now to create our required expressions. So AP-dash is equal to a divided by a plus b times A B-dash. And we know that A B-dash is the horizontal distance from the point A to the point B so that's just the difference between the x-coordinates of those two points, that is x2 minus x1. And tidying up, we get a single fraction - ax squared minus ax... Sorry, ax2 minus ax1 all divided by a plus b. Now, we're not just looking for that distance, we're in fact looking for the x-coordinate of the point P. And we know that the x-coordinate of the point P is just the x-coordinate of the point A plus this horizontal distance between A and P that we've just worked out above. And so we get this expression.

[x equals x1 plus A P-dash. equals x1 plus ax2 minus ax1 all over a plus b. Equals x1 plus ax2 minus ax1 all over a plus b. Equals ax1 plus bx1 all over a plus b, plus ax2 minus ax1 all over a plus b. Equals bx1 plus ax2 all over a plus b, as required.]

NARRATOR: And we can now create common denominators, put the fraction together to find that the x-coordinate of the point P is b times x1 plus a times x2 all divided by a plus b as was required in the question. We can now use similar logic for the y-coordinate of the point P. And we note that PP-dash, that is the vertical distance from A to P, is equal to a divided by a plus b times BB-dash, which is the vertical distance from A to B. And we know that this vertical distance from A to B is simply the difference in the y-coordinates between those two points, so that is y2 minus y1, which simplifies as a single fraction to a times y2 minus a times y1 all divided by a plus b. Again, though, the y-coordinate of the point P does not equal this expression in itself. The y-coordinate of the point P would be equivalent to the y-coordinate of the point A plus this new vertical distance that we've established above.

[Equations: y equals y1 plus P P-dash. Equals y1 plus ay2 minus ay1 all over a plus b. Equals ay1 plus by1 all over a plus b, plus ay2 minus ay1 all over a plus b. Equals by1 plus ay2 all over a plus b, as required.]

NARRATOR: And again, creating a common denominator and putting those two fractions together, we find that the y-coordinate of the point P, that is y, is equal to b times y1 plus a times y2 all divided by a plus b, as was stated in the question. And so we have shown that the coordinates of the point P are as stated.