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Copyright 2004, Rick Bowman

Conquering Coordinate Geometry

Part One

plot points on anx-ycoordinate (also called

Cartesian) system

calculate the gradient (slope) of a line connecting 2 points

calculate the distancebetween two points

determine the midpoint of two points

In this section you will learn how to

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Part 1

Plotting and

Naming Points

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Look at thisx-y system at

right

The centre is the

ORIGIN O

O

Positivedirections are:

RIGHT +UP

+

Negative directions are:

LEFT

DOWN

This line is called the

x-axis

This line is called they-axis

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

How do we name points

anywhere on the graph?

All points are identified by two

COORDINATES, anx-coordinate

and ay-coordinate in the format (x,y)

E.g. consider the point

(+3, 4)

Begin first at O

O

The +3 tells you to go 3

units in the positive x-direction (right)From where you are now,

the -4 tells you to go 4

units in the negative y-direction (down)

(3,4)

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Here is a portion of anx-y system

Lets graph some points. Each

time, start at O (the origin)

(3, 2)(3, 2)

(1, 4)

(1, 4)

(0, 1)(0, 1)

(3,2)(3,2)

Note that the first number (x)in each

ordered pair tells you left/right.

O

Move right3 then up 2

Move left1 then up 4Move right/left0 then up 1Move right3 then down 2

and the second number (y) in each

ordered pair tells you up/down

(4,2 ) Move left4 then down 2(4,2)

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

If a point is situated on thex-axis, then its 2ndnumber (they-

coordinate) is ZERO

This is because there is no

upward or downward

movement needed to get

to the point from theorigin (where you start

from)

KEY FACT

Do you know why?

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KEY FACT

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

If a point is situated on they-axis, then its 1st number (thex-

coordinate) is ZERO

This is because there is no

right or left movement

needed to get to the point

from the origin

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Now try

Worksheet #1

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Part 2

Gradient (slope) of a

l ine joining 2 points

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Gradient (slope)measures the steepness of a line

is positive if the line leans to the right

is negative if the line leans to the left

is zero if the line is horizontal

is not defined if the line is vertical

m> 0

m< 0

m= 0

mnot defined

KEY FACTSis defined as

RunHorizontal

RiseVerticalm

RISERUN

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Here are some examples. Study them closely!!

4 cm

2 cm2

2

4m

1

3

3

1m

Run

Risem

8

8

18

8 m

2

3

3

2mm = 0

5 km

3 km

3

5m

Leans left, som is negative

NOTE Sometimes Rise and Run will have units (cm, m, km) butthe radient never has units.

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Comparing gradient

The gradients of the following 7 lines are + , +2, +1,

0, 1,2, (not in order). Can you match the

correct line with its gradient? (answers next slide).

(a) (b) (c)

(d) (e) (f)

(g)

Remember the larger the size of the gradients value (ignoring the+/- sign), the steeper the line A line with gradient of 3 is steeperthan one of +2 which in turn is steeper than one of 1 or 1. Thesign just tells you which way it leans.

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Did you match correctly?

(a) (b) (c)

(d) (e) (f)

(g)

m= +2 m=2

m= 0

m=1m= +

m=m= + 1

Did you remember to make the gradient negativeif the

line leans backwards? (Like in (b), (d) and (f)) ??

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y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.

A(4,3)

B(1,3)

On the graph we have

two points A(4,3) and

B(1,3)Our aim is to find the

rise, run and gradient of

the line joining them.

First, join AB and

extend line both ways

Now make a triangle ABC

using A & B as vertices

C

This must be a right-angled triangle andthe line in question (AB in this case)should be the hypotenuse.

Finding gradient between points

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

B(1,3)

A(4,3)

C

To determine the RUN,

measure the horizontal

distance BCRun = 3 units

3

To determine the RISE,

measure the vertical

distance CA. 6

Rise = 6 units

To determine the

GRADIENTwork out

RISE RUN.

Gradient = 6 3 = 2

Finally, the blue line leans RIGHT so gradient is +2.

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

A(4,3)

B(1,3)

Note that you could have drawn

the triangle differently, with C in

the opposite position to where it

was before.newC

oldC

Regardless of which position

you choose for C (either aboveor below the blue line),

the rise and run will still be

the same

Rise BC will still be 6 and run

CA will still be 3.

3

6

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Example....Find the gradient

of the line joining A(3, 2)

to B(4,3)

Step 1Plot and join A, B.

A(3,2)

B(4,3

Step 2Make ABC. You

can put C in either of 2

positions (marked with ).

Lets pick the top right.

C

Step 3Measure rise & run

Rise is 5

5

Run is 7

7

Step 4Determine which way the line leans. Since it leans

LEFT, the gradient will be NEGATIVE, som= 5/7

Key point: Its easier not to bother introducing negatives until right at the endwhen you consider which way the line leans.

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x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

To work out the gradient of a line,

it doesnt matter which 2 points

you use, as long as they both lie on

the line.

To illustrate this idea, plot

the points A(1,3), B(0,1)

and C(2, 3)

A(1,3)

B(0,1)

C(2, 3)

They all lie on the same line

KEY FACT

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Lets choose A and C as our 2 points.

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

A(1,3)

C(2, 3)Make a triangle and work outrise, run and gradient:

m=

Run

Rise

3

6 = 2

3

6

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Now suppose instead we chose B and

C as our 2 points.

Make a triangle and work out

rise, run and gradient:

m=

Run

Rise

2

4

= 2 !x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

B(0,1)

C(2, 3)2

4

......and what do you think would

happen if we chose A and B as

another option? A(1,3)

This timeRun

Rise

1

2 = 2 !

2

1

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Now try

Worksheet #2

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You can also calculate the gradient

without drawing a graph.

Suppose we graph two points A(1, 2) and B(3, 6). (Not to scale!)

x

y

A(1,2)

B(3,6)

Did you know thex andy

coordinates tell you the height and

distance of each point relative to

the axes?

1 3

2

6

It is now just a matter of

subtracting these to get the rise

and run...........

(31 )

(6

2)

RUNRise = 62 = 4

So m=Run

Rise

2

4 = 2

Run = 31 = 2

2

4

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This can be extended to any two points.So we use algebra to generalise.

Suppose we have two points A(x1,y1) and B(x2,y2).

x

y

A(x1,y1)

B(x2,y2)

As on the previous slide, thex and

ycoordinates tell you the height

and distance of each point relative

to the axes.

x1 x2

y1

y2

It is now just a matter of

subtracting these to get the rise

and run...........

(x2x1)

(y2y1)

RUNRise = y2y1Run = x2x1

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So the gradient, mis given by

12

12

xx

yym

Run

Risem

or, using the fact that the RISE is (y2

y1

)

and the RUN

is (x2x1)as shown on the previous slide..........

The Gradient formula.....

Run (x2x1)

Rise(y

2y1)

KEY FACT

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Example..

Find the gradient of the

line joining (3, 5) to (1,1 )

Step 1 Call x1= 3, y1= 5 and x2= 1, y2=1.

You could have labelled these the other way hadyou preferred, i.e. x2= 3, y2 = 5 and x1= 1, y1= 1just as long as you dont mix them upStep 2 Write out the gradient formula

12

12

xx

yym

Step 3 Replace the letters in the formula with thevalues you gave them in Step 1

12

12

xx

yym

31

51

2

6

= 3

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We could have done this using a graph

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Step 1 Plot & join A(3, 5)

to B(1,1)

(3,5)

(1,1)

Step 2 Make the triangle as

shown on Slide 14

Step 3 Visually measure

rise and run

Rise = 6

6

Run = 2

2

Step 4 Apply ruleRun

Risem

i.e. m=+3

and remembering line leans right

means gradient is positive

2

6m

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Now try

Worksheet #3

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Distance between two

points

Part 3

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Essential previous knowledge

PYTHAGORASS THEOREM

In any right angled triangle

ABC, whose side lengths are

a, b and c

a

b

c

The hypotenuse, c is given by

c

2

= a2

+ b2

, or..

22 bac This is PythagorassTheorem and is the basisfor the Distance Formulato come on the next slide

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b

a

c

In previous slides we

called side lengths a

and b by the names

run and rise.

rise

run

d

So we can change

Pythagorass theorem from

to

d= 22 )()( riserun

Now from Slide 23,

Rise = (y2y1)& Run = (x2x1)

22bac

(x2x1)

(y2y1)

so d=2

122

12 )()( yyxx

This is called the DISTANCE FORMULA

Developing the DISTANCE FORMULA

Remember this??

KEY FACT

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Example..

Find the distance between the points (3, 2) and (3,6)

using ALGEBRA

Step 1 Call x1=3, y1= 2 and x2= 3, y2=6

REMEMBER you could have interchanged these and called x2 = -3,y2= 2 and x1= 3, y1= -6. But youre NOT ALLOWED to mix themStep 2 substitute x1=3, y1= 2, x2= 3, y2=6 into

212

212 )()( yyxxd

22 )26()33( d

i.e.22)8()6( d1006436 di.e.

to get

d= 10

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Example..

Find the distance between the points (3, 2) and (3,6)

using A GRAPH

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

Step 1 Plot points, join andmake triangle

(3,6)

(3, 2)

Step 2 Find values of riseand run from thegraph. Label 6 and 8

6

8Step 3 Use Pythagoras.d

d2= 62+ 82

d2= 36 + 64

d2= 100

d= 10

10

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Now try

Worksheet #4

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Part 4

The M idpoint of two

points

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Background..

Consider the two numbers 6 and 10.

The number exactly halfway between them ,

in other words, their MIDPOINT, is 8.

In this case, the midpoint can easily be worked

out by counting inwards from 6 and 10, but you

need to know the midpoint can also be found by

averaging the two numbers. KEY FACT8

2

106

This averaging is a really useful process when the

numbers are not as easy to work with as 6 and 10

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Example

Find the midpoint of7 and 12.

Counting inwards is fairly tedious and

time consuming. Its quicker and easier to

work out their average.

2

127 M

2

12

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This concept can now be used to find the midpoint of

two points on an x-y graph.

Example. Find the midpoint of (3,4) and (1, 2)

The trick here is to use our averaging method, first on the 3 and 1(the two xvalues) and then on the 4 and 2 (the two yvalues)

x

y

-5.0

-4

-3

-2

-1

1

2

3

4

5.0

-5.0 -4 -3 -2 -1 1 2 3 4 5.0

(3,4)

(1,2)Step 1 Average the xs

12

13

Step 2 Average the ys

12

24

Step 3 M = (1,1)

(1,1)

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We can now come up with a general formula that

finds the midpoint of any two points (x1,y1) and

(x2,y2)

Remembering that to get the midpoint we

average the xs then average the ys, our

formula is

2

,2

2121 yyxxM

A word of caution The midpoint is a POINT and soshould be written as a point, ie with brackets, and twonumbers inside, separated by a comma

KEY FACT

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Now try

Worksheet #5

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But f irst, an essential background ski l l

Changing the subject of a formula

Suppose you begin with an equation like 2x+ y= 5

or 4x5y= 7,

You need to be able to rearrange this so that it

begins with y = , in other words, with yas the

subject

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Example 1

Make ythe subject of 2x+ y = 5.

We need to get rid of the 2x soywill be left on its own.As 2xis linked to the yby a +sign, we UNDO this bysubtracting2xfrom both sides.2x+ y2x= 52x

Simplify the left side, and the 2x disappears as planned.y= 52x or y =2x + 5

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Example 2

Make ythe subject of 3y = x2.

We need to get rid of the 3soywill be left on its own.As 3 is linked to the yby a sign, we UNDO this bydividingboth sides by 3.

3

2

3

3

xy

Simplify the left side, and the 3 disappears as planned.

3

2x

y

The right hand side needs to be changed as follows..

3

2x

y SPLIT3

2

3

xy REWRITE

3

2

3

1 xy

It is better to getthe xalone, witha number infront.

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Example 3

Make ythe subject of 4y + 8x = 2.

First we need to get rid of the 8x so4y will be left onits own. As 8xis linked to the 4yby a + sign, we UNDOthis by subtracting8xfrom both sides4y+ 8x8x =28x

Simplify the left side, and the 8xdisappears as planned.4y=28x

Second we need to get rid of the 4 soywill be left onits own. As 4 is linked to the yby a sign, we UNDOthis by dividingboth sides by 44

82

4

4 xy

Simplify the left side, and the 4 disappears as planned.

4

82 xy

SPLIT4

8

4

2 xy

xy 22

1

2

1

2

xy

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Now try

Worksheet #6