WEBVTT

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Hi, my name is Richard.

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We're looking at TMUA 2021, paper two,
question 7.

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Here it is, so let's read it out.

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We've got a circle has equation X - 9 ^2
+ y + 2 ^2 = 4.

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We have a square with vertices at 1012
minus 1/2 and -1 zero.

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We are told that a straight line bisects
both the area of the circle and the area

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of the square.

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And we're asked what is the X coordinate
of the point where that straight line

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meets the X axis?

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OK, before we get into the problem,
let's think about straight lines

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bisecting circles and squares.

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So if we start with a circle,
we'll pick out the centre.

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Here we can easily imagine that the only
straight lines which are going to bisect

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that circle into two equal areas would be
lines which pass through the centre.

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So this line clearly creates 2 equal
areas passing through the centre,

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as does this line.

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But you can imagine if the line does not
pass through the centre,

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we do not get that bisection into two
equal areas.

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OK,
so that's the key point for the circle.

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And if we now move on to a square,
let's imagine we've got a nice square

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here with horizontal and vertical sides.

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Again,
we can think about the centre of the

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square,
which would be where the two diagonals

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cross.

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And again,
in order to have a line which bisects

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that square into two equal areas,
our line must pass through the centre.

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There are four kind of obvious examples
of that,

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like this vertical line and this
horizontal line creating 2 equal areas in

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both cases.

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And also these diagonals,
clearly they divide our square into two

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equal areas.

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But also,
if you imagine any other line which

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passes through the centre,
perhaps you can imagine this line as a

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rotation of the vertical one.

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And because these two areas are the same,
that line too will have bisected our

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square into two equal areas.

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So actually it's the same for both shapes.

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For the circle we need a line which goes
through the centre,

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and also for the square we need a line
which goes through the centre.

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OK,
now let's move on to consider the actual

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circle and the square that we have in our
question.

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So the circle,
we can see its centre has coordinates 9 -,

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2 reading off from the equation.

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So that might be somewhere over here.

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We can also see the radius is 2,
the square root of this right hand side.

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So actually if we wanted to imagine it,
we would need to imagine it something

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like that with a radius of two touching
the X axis, the square,

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Let's plot its vertices.

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So we've got one at 10,
we've got another one at 122.

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Above that we've got -1 two,
so that'll be two to the left and also -1

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zero.

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So that will be here.

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So we can imagine our square looks like
this,

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and hopefully we can imagine that the
centre of that square where the two

68be9dd5-168a-4c35-8ca8-adda93109001-2
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diagonals cross, this would be the .01.

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OK.

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So we know that in order to bisect both
of these,

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we need a straight line which passes
through both of these points.

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So we can now imagine that this straight
line looks like this,

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and we have to work out the X coordinate
of where this straight line meets the X

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axis.

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So we're after this value here on the X
axis.

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So I'm going to work out the equation of
the straight line and use that to work

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out that X coordinate.

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I'm going to start with the gradient.

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So I know these two points on the line as
I move from here to here,

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I can see my change in Y is from one down
to -2.

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So my gradient calculation,
I have a change in Y of -3.

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I have a change in X of +9 going from
zero to 9.

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So the gradient of that line is -1 third.

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And now I'm going to use the standard
equation for a straight line like this.

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Y -, y one equals MX minus X1,
where M is the gradient and X1 Y 1 is a

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known point on the line.

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So I'm going to use my known point as 01
here, which means X1 is going to be 0,

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Y 1 is going to be 1,
and M is going to be -1 third.

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So this gives me y - 1 = - 1/3 X -0 or y
- 1 = - 1/3 X.

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And to work out where that line passes
through the X axis,

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I need to set Y to be 0.

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So setting Y as zero gives me -1 is minus
1/3 X and now multiplying through by -3

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will give me the X = 3.

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So I know the X1 at this point is 3.

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So I look over at my possible answers.

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I'm looking for three and there it is as
B.