Text description - screencast for exercise 14

[The narrator reads out the onscreen text.]

NARRATOR: Exercise 14. Let M be the point (2, 3) and let l be a line through M which meets 2x plus y minus 3 equals 0 at A and meets 3x minus 2y plus 1 equals 0 at B. If M is the midpoint of AB, find the equation of l.

[On an x and y graph, a blue line marked 3x minus 2y plus 1 equals 0 runs diagonally upwards, crossing the x-axis, then the y-axis. A black line marked 2x plus y minus 3 equals 0 runs diagonally downwards crossing the y-axis, then the x-axis and then the line l. The red line l runs diagonally upwards, crossing the y-axis then the x-axis.]

NARRATOR: It can help to first consider this situation diagrammatically, that is that A is a point in the line 2x plus y minus 3 equals 0 and B is a point in the line 3x minus 2y plus 1 equals 0, and M is the midpoint of A and B with coordinates (2, 3). The line passing through A, M and B is called l, and it is the equation of this line that we wish to establish. Let a be the x-coordinate of the point A, remembering that a is a point on the line 2x plus y minus 3 equals 0.

[When x equals a: 2a plus y minus 3 equals 0. y equals 3 minus 2a.]

So when x equals a, y is equal to 3 minus 2a and therefore the coordinates of the point A can be expressed as (a, 3 minus 2a). Let b be the x-coordinate of the point B, remembering that B is a point on the line 3x minus 2y plus 1 equals 0.

[When x equals b: 3b minus 2y plus 1 equals 0. 2y equals 3b plus 1. y equals 3b plus 1 all over 2.]

NARRATOR: When x equals b, y is equal to 3b plus 1 all divided by 2. And so the coordinates of B can be expressed as (b, 3b plus 1 all divided by 2). Now, we know that M, with coordinates (2, 3), is the midpoint of A and B for which we've just established expressions for the coordinates. We know that the midpoint of A and B can be found by averaging the x-coordinates of A and B and averaging the y-coordinates of A and B. So that is the x-coordinate is equal to a plus b divided by 2, which would mean that that is equal to 2. And the b-coord...sorry, the y-coordinate of the midpoint of AB is found by averaging the y-coordinates of A and B. That is 3 minus 2a plus 3b plus 1 over 2 all divided by 2, which therefore must equal 3, the y-coordinate of the midpoint. And so we gain these two equations.

[a plus b all over 2 equals 2. 3 minus 2a, plus 3b plus 1 all divided by two, all over 2 equals 3.]

NARRATOR: Considering each of these two equations, the first equation, that is the one relating the x-coordinates, allows us to create a simple equation relating a and b.

[a plus b all divided by 2 equals 2. a plus b equals 4. b equals 4 minus a.]

NARRATOR: We'll make b the subject, making it b equals 4 minus a. In the second equation, relating the y-coordinates, we can simplify this equation by first multiplying everything by 2 and then by 2 again to eliminate the fractions, and then collecting like terms to get the equation negative 4a plus 3b equals 5.

[3 minus 2a, plus 3b plus 1 all divided by two, all over 2 equals 3.
3 minus 2a plus 3b plus 1 all over 2 equals 6. 6 minus 4a plus 3b plus 1 equals 12. negative 4a plus 3b equals 5.]

NARRATOR: So substituting b equals 4 minus a into the second equation we get that a equals 1.

[Negative 4a plus 3 times 4 minus a equals 5. Negative 4a plus 12 minus 3a equals 5. Negative 7a equals negative 7. a equals 1. b equals 4 minus 1 equals 3.]

NARRATOR: And substituting a equals 1 to find b, we find that b is equal to 3. So if a equals 1 and b equals 3, we can determine the coordinates of A and B, that is that point A has coordinates (1, 1), and that point B has coordinates (3, 5). So now we know that the line l passes through the points A with coordinates (1, 1), M with coordinates (2, 3) and B with coordinates (3, 5). And any combination of these points can be used to find the equation of l. We start by finding the gradient of the line l. And that's found by looking at the difference between the y-coordinates divided by the difference between the x-coordinates. In this case we're going to use the points A and M. So we're looking at 3 minus 1 divided by 2 minus 1 which gives 2 over 1, or a gradient of 2. And using y minus y1 equals M times x minus x1 along with our gradient of 2 and the point A with coordinates (1, 1), we can establish the equation for the line l.

[y minus 1 equals 2 times x minus 1. y minus 1 equals 2x minus 2. y equals 2x minus 1.]

NARRATOR: And so therefore the equation of l is y equals 2x minus 1.