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.
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.
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, 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.
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.
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.
So substituting b equals 4 minus a
into the second equation
we get that a equals 1.
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.
And so therefore the equation of l is y equals 2x minus 1.