WEBVTT

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Hi, I'm Fiona.

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Let's have a look at TAMIMA 2021 Paper 2
and question 2.

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The coordinate points A and C are
opposite vertices of the square ABCD.

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And we're asked what is the equation of
the straight line through B&

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D and then given options A to F to choose
from at the end.

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So let's have a look at drawing a picture
of what's going on here.

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So I'm just going to draw a generic
square.

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This is not the correct orientation of
this square at all,

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but just to give myself an idea of what's
going on,

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I've got vertices ABC and D Vertex A is
02 and vertex C is 40.

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So here's a picture to give me an idea of
what's going on,

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and I'm asked what is the equation of the
straight line through B&D?

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Now because the diagonals of a square
bisect one another perpendicularly and

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also in the centre of the square,
I can see that the equation of the line

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that goes through vertex B and through
vertex D is the perpendicular bisector of

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the line segment that joins the vertex A
to vertex C.

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And when I want to find a line,
I either need 2 points on that line or I

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need a gradient and a coordinate point.

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And so for this one, the coordinate,
A coordinate point that lies on this line

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is the midpoint of this line segment
between A and C And I am going to be able

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to find the gradient by calculating the
gradient of the line segment between A

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and C and then getting the negative
reciprocal of that.

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So let's start with the gradient.

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So the gradient of the line segment
between A and C will be 2 - 0.

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So that's 2 / 0 -, 4.

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So that's 2 / -, 4.

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And that's going to be minus 1/2.

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So then the gradient that I'm looking for
of the, sorry,

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the gradient of the line I'm looking for,
I'm going to call that M.

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It's going to be the negative reciprocal,
so that's going to be a +2.

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Next, I want to find the midpoint of AC,
which is a point that lies on the line

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that I'm looking for.

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And the midpoint of any 2 coordinate
points is just the average of their

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coordinates.

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So the midpoint between A and C is going
to be the average of the X ordinates,

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which is 4 + 0 / 2, so that's two,
and the average of the Y ordinates,

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which is 0 + 2 / 2,
so that's going to be 1.

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Now I've got a point that lies on the
line I'm looking for.

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I've got its gradient,
so I'm going to use my formula,

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which is Y minus Y 1 = m on X -, X one.

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So Y 1 is going to be 1 so I'll have y -,
1 and then M is going to be two and X1 is

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going to be two as well.

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Now just tidying this up,
I have y - 1 = 2 X -4 and what form do I

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want my line to be in?

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I want Y equals and I want a X term and
then my constant term.

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So I've got y = 2 X and then -4 + 1 is
going to be -3.

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So that's the line I'm looking for,
and I can see that that is option E So E

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is the answer to this question.