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NARRATOR: Exercise 14.

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Let M be the point (2, 3)

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and let l be a line through M which
meets 2x plus y minus 3 equals 0 at A

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and meets 3x minus 2y plus 1
equals 0 at B.

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If M is the midpoint of AB,
find the equation of l.

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It can help to first consider
this situation diagrammatically,

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that is that A is a point in the line
2x plus y minus 3 equals 0

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and B is a point in the line
3x minus 2y plus 1 equals 0,

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and M is the midpoint of A and B
with coordinates (2, 3).

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The line passing through A, M and B
is called l,

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and it is the equation of this line
that we wish to establish.

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Let a be the x-coordinate
of the point A,

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remembering that a is a point on
the line 2x plus y minus 3 equals 0.

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So when x equals a,
y is equal to 3 minus 2a

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and therefore the coordinates
of the point A can be expressed

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as (a, 3 minus 2a).

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Let b be the x-coordinate
of the point B,

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remembering that B is a point on the
line 3x minus 2y plus 1 equals 0.

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When x equals b, y is equal to
3b plus 1 all divided by 2.

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And so the coordinates of B
can be expressed

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as (b, 3b plus 1 all divided by 2).

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Now, we know that M,
with coordinates (2, 3),

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is the midpoint of A and B

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for which we've just established
expressions for the coordinates.

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We know that the midpoint of A and B
can be found

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by averaging the x-coordinates
of A and B

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and averaging the y-coordinates
of A and B.

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So that is the x-coordinate
is equal to a plus b divided by 2,

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which would mean that
that is equal to 2.

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And the b-coord...sorry, the
y-coordinate of the midpoint of AB

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is found by averaging
the y-coordinates of A and B.

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That is 3 minus 2a plus
3b plus 1 over 2

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all divided by 2,

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which therefore must equal 3,
the y-coordinate of the midpoint.

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And so we gain these two equations.

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Considering each
of these two equations,

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the first equation, that is
the one relating the x-coordinates,

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allows us to create a simple equation
relating a and b.

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We'll make b the subject,
making it b equals 4 minus a.

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In the second equation,
relating the y-coordinates,

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we can simplify this equation
by first multiplying everything by 2

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and then by 2 again
to eliminate the fractions,

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and then collecting like terms

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to get the equation
negative 4a plus 3b equals 5.

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So substituting b equals 4 minus a

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into the second equation

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we get that a equals 1.

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And substituting

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a equals 1 to find b,

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we find that b is equal to 3.

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So if a equals 1 and b equals 3,

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we can determine the coordinates
of A and B,

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that is that point A
has coordinates (1, 1),

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and that point B
has coordinates (3, 5).

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So now we know that the line l
passes through

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the points A with coordinates (1, 1),
M with coordinates (2, 3)

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and B with coordinates (3, 5).

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And any combination of these points

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can be used to find
the equation of l.

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We start by finding
the gradient of the line l.

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And that's found by looking at

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the difference between
the y-coordinates

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divided by the difference
between the x-coordinates.

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In this case we're going to use
the points A and M.

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So we're looking at 3 minus 1
divided by 2 minus 1

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which gives 2 over 1,
or a gradient of 2.

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And using y minus y1
equals M times x minus x1

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along with our gradient of 2 and
the point A with coordinates (1, 1),

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we can establish
the equation for the line l.

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