conq part 1
TRANSCRIPT
-
8/13/2019 Conq Part 1
1/45
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
-
8/13/2019 Conq Part 1
2/45
Part 1
Plotting and
Naming Points
-
8/13/2019 Conq Part 1
3/45
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
-
8/13/2019 Conq Part 1
4/45
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)
-
8/13/2019 Conq Part 1
5/45
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)
-
8/13/2019 Conq Part 1
6/45
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?
-
8/13/2019 Conq Part 1
7/45
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
-
8/13/2019 Conq Part 1
8/45
Now try
Worksheet #1
-
8/13/2019 Conq Part 1
9/45
Part 2
Gradient (slope) of a
l ine joining 2 points
-
8/13/2019 Conq Part 1
10/45
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
-
8/13/2019 Conq Part 1
11/45
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.
-
8/13/2019 Conq Part 1
12/45
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.
-
8/13/2019 Conq Part 1
13/45
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)) ??
-
8/13/2019 Conq Part 1
14/45
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
-
8/13/2019 Conq Part 1
15/45
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.
-
8/13/2019 Conq Part 1
16/45
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
-
8/13/2019 Conq Part 1
17/45
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.
-
8/13/2019 Conq Part 1
18/45
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
-
8/13/2019 Conq Part 1
19/45
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
-
8/13/2019 Conq Part 1
20/45
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
-
8/13/2019 Conq Part 1
21/45
Now try
Worksheet #2
-
8/13/2019 Conq Part 1
22/45
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
-
8/13/2019 Conq Part 1
23/45
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
-
8/13/2019 Conq Part 1
24/45
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
-
8/13/2019 Conq Part 1
25/45
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
-
8/13/2019 Conq Part 1
26/45
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
-
8/13/2019 Conq Part 1
27/45
Now try
Worksheet #3
-
8/13/2019 Conq Part 1
28/45
Distance between two
points
Part 3
-
8/13/2019 Conq Part 1
29/45
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
-
8/13/2019 Conq Part 1
30/45
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
-
8/13/2019 Conq Part 1
31/45
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
-
8/13/2019 Conq Part 1
32/45
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
-
8/13/2019 Conq Part 1
33/45
Now try
Worksheet #4
-
8/13/2019 Conq Part 1
34/45
Part 4
The M idpoint of two
points
-
8/13/2019 Conq Part 1
35/45
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
-
8/13/2019 Conq Part 1
36/45
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
-
8/13/2019 Conq Part 1
37/45
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)
-
8/13/2019 Conq Part 1
38/45
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
-
8/13/2019 Conq Part 1
39/45
Now try
Worksheet #5
-
8/13/2019 Conq Part 1
40/45
-
8/13/2019 Conq Part 1
41/45
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
-
8/13/2019 Conq Part 1
42/45
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
-
8/13/2019 Conq Part 1
43/45
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.
-
8/13/2019 Conq Part 1
44/45
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
-
8/13/2019 Conq Part 1
45/45
Now try
Worksheet #6