6 mat unit 4 coordinate geometry
TRANSCRIPT
11
Unit 4 Coordinate GeometryUnit 4 Coordinate Geometry
Lesson 1: Midpoint of a line and distance between 2 points
Lesson 2: To find the gradient of a line
Lesson 3: Gradient of a Lesson 3: Gradient of a line making an angle line making an angle with the axes with the axes
Lesson 4: Rearranging Lesson 4: Rearranging the equation of a the equation of a straight linestraight line
Lesson 5/6: Finding the Lesson 5/6: Finding the equation of lines from equation of lines from given informationgiven information
Lesson 7: Parallel Lines Lesson 7: Parallel Lines
Lesson 8: Perpendicular Lesson 8: Perpendicular Lines Lines
Lesson 9: Collinear Lesson 9: Collinear points and drawing linespoints and drawing lines
Lesson 10:Intercept – Lesson 10:Intercept – intercept methodintercept method
Lesson 11: Lesson 11: Simultaneous EquationsSimultaneous Equations
Lesson 12/13: Lesson 12/13: Simultaneous Equations- Simultaneous Equations- SubstitutionSubstitution
Lesson 14/15: Lesson 14/15: ApplicationsApplications
Starters
22
Co-ordinate Co-ordinate GeometryGeometry
Lesson 1Lesson 1Midpoint of a line and Midpoint of a line and
distance between 2 pointsdistance between 2 points
33
y
x
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
2
2
4
4
6
6
8
8
10
10
– 2
– 2
1 2 1 2( ) ( ),
2 2
ie we average the points
x x y ymp
(x1,y1)
To find the mid point of a line
A line segment is part of a line
The mid point is exactly half way betweenthe end points
In the example
1 1 2 2
1 2 1 2
, ( 4,0), , 0,8
( ) ( ),
2 2
( 4 0) (0 8),
2 2
2,4
x y x y
x x y ymp
mp
mp
(x2,y2)
Midpoint of a lineMidpoint of a line
44
2,
2
)( 2121 yyxx
The midpoint of the line joining two points (x1,y1) and (x2,y2) is
.
1 1( , )x y
Midpoint of a Line
55
Be careful when there are negatives
2, 2mp
1 1 2 2, (6, 1), , 2, 3x y x y 1x 1y 2x 2y
It is useful to put the correct
coordinate over the top
then you won’t get confused
1 2 1 2( ) ( ),
2 2
x x y ymp
(6 2) ( 1 3)
,2 2
mp
(6 2) ( 1 3),
2 2mp
4 ( 4),
2 2mp
66
Sometimes you are given the midpointMid point of AB is (4,5) If A is (2,9) find B
1 1 2 2, (2,9), , ( , ), 4,5A x y B x y x y mp
(2 ) (9 )(4,5) ,
2 2
x y
(2 ) (9 )4 , 5
2 2
x y
8 2 , 10 9x y
6, 1x y
(6,1)B
77
Calculate the midpoint of the lines joining the following pairs of points;
1. (9 , 4) and (6 , 2)
2. (3 , 7) and (-4 , -1)
3. (-1 , 5) and (0 , -4)
4. (-5 , 7) and (5 , -7)
5. (-2 , -4) and (3 , -6)
6. (3.7 , -1.8) and (-2.4 , 1.5)
7. (x , 2x) and (-x , -2x)
8. (p , 0) and (-p , -p)
9. (z,- z) and (-z, z)3 3
88
Ex 2.1: #1a-I (1column) 2-5Ex 2.1: #1a-I (1column) 2-5WB P170#12-25 WB P170#12-25
(15mins) (15mins)
99
2 2 2
2 2 22 1 2 1
2 22 1 2 1
( ) ( )
( ) ( )
h a b
d x x y y
d x x y y
y
x
1
1
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
x
x
(x1,y1)
(x2,y2)
We can find the distance between two points by Pythagoras
Distance between two points
1 1 2 2( , ) ( 8,2) ( , ) ( 2,6)x y x y 1x 1y 2x 2y
(y2-y1)
(x2-x1)
1010
2 2 2
2 2 22 1 2 1
2 22 1 2 1
( ) ( )
( ) ( )
h a b
d x x y y
d x x y y
y
x
1
1
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
x
x
(x1,y1)
(x2,y2)
We can find the distance between two points by Pythagoras
7.2(1 )d dp units
Distance between two points
1 1 2 2( , ) ( 8,2) ( , ) ( 2,6)x y x y 1x 1y 2x 2y
2 22 1 2 1( ) ( )d x x y y
2 2(( 2) ( 8)) (6 2)d 2 2( 2 8) (4)d
2 2(6) (4)d
36 16d
52d
(y2-y1)
(x2-x1)
1111
Calculate the distance between the following points;
1. (9 , 4) and (6 , 2)
2. (3 , 7) and (-4 , -1)
3. (-1 , 5) and (0 , -4)
4. (-5 , 7) and (5 , -7)
5. (-2 , -4) and (3 , -6)
6. (3.7 , -1.8) and (-2.4 , 1.5)
7. (x , 2x) and (-x , -2x)
8. (p , 0) and (-p , -p)
9. (z,- z) and (-z, z)3 3
= 3.6 units
= 10.6 units
= 9.1 units
= 17.2 units
= 5.4 units
= 6.5 units
20
5
= 4z units
x units
p units
1212
Remember
3
4
5 13
5
12
25
24
7
1313
End of Lesson 1End of Lesson 1
Ex 2.2: #a-j odd, 2,3,4,5,7,8,9Ex 2.2: #a-j odd, 2,3,4,5,7,8,9WB P168 #1-11WB P168 #1-11
1414
Lesson 2Lesson 2
To find the gradient of a To find the gradient of a lineline
1515
2 1
2 1
Gradient
( )
( )
m
ym
xy y
mx x
y
x
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
2
2
4
4
6
6
8
8
10
10
– 2
– 2
(x1,y1)
(x2,y2)
In the example
To find the gradient of a lineTo find the gradient of a line
1 1 2 2, ( 4,0), , 0,8x y x y 1x 1y 2x 2y
(8 0)
(0 ( 4))m
8
42
m
m
1616
Sometimes we have unknowns
Be careful when there are negatives
1 1 2 2, ( 9, 3), , 1, 4x y x y 1x 1y 2x 2y
1
8m
2 1
2 1
( )
( )
y ym
x x
(( 4) ( 3))
(( 1) ( 9))m
( 4 3)
( 1 9)m
The gradient of the line between
(4,2) and ( ,5) is - 6. Find p p p
(5 2)6
( 4)p
p
( 4)( 6) 3p p 2 10 24 3p p 2 10 21 0p p
( 7)( 3) 0p p 7, 3p p
1717
End of Lesson 2End of Lesson 2
WB P174 #26-35Ex 2.3: # 2-9
1818
Lesson 3Lesson 3
Gradient of a line making an angle with the axes
1919
y
x
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
2
2
4
4
6
6
8
8
10
10
– 2
– 2
(x1,y1)
(x2,y2)
In the example
63.4 (1 )dp
To find the Gradient of a line making an angle θ with the x axis
2 1
2 1
y ym
x x
tanO
A Gradient =
y
x
O
Atan m
1 1 2 2, ( 4,0), , 0,8x y x y 1x 1y 2x 2y
(8 0)
(0 ( 4))Tan
8
4Tan
2Tan 1(2)Tan
O
A
2 1
2 1
tany y
x x
2020
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
2
2
4
4
6
6
8
8
10
10
– 2
– 2
36.9 143.1
eg find the angle a line with the gradient of – ¾ makes with the positive direction of the x axis
3
4Tan
1 3tan
4
36.9
smallest angle is 36.9°
Largest angle is 36.9 180 143.1
2121
1
tan
tan (shift tan 9999999)
90
gradient undefined
Eg Ex 2.4
60
Note
1. Calculate the gradient of a line making the following
angle with the positive direction of the x-axis
0a) 24
tan 24 0.4452 gradient 0b) 126
tan126 1.376 gradient
3. Calculate angle line makes with positve x axis
given gradient is 3
3gradient
tan 3 1tan 3
2222
End of Lesson 3End of Lesson 3
Ex 2.4: #1 odd, 2 all, 3 odd, 4 Ex 2.4: #1 odd, 2 all, 3 odd, 4 -7-7
2323
Lesson 4Lesson 4
Rearranging the equation of a straight
line
2424
If a line is written in the form y mx c
3we can now see the gradient is and the intercept is -1
2y
Eg Ex 2.5 #1Find the gradient of
Rearranging the equation of a straight line
We need to make y the subject
There are two forms for straight lines
-interceptyGradient
0ax by c
and
y mx c 2 3 2 0y x 2 3 2y x 2 3 2
2 2 2y x ( all of both sides by 2)
3= 1
2y x
2 3 2 0y x
2525
0
( tells us angle angle with Positive direction
of the x axis going counter clockwise is 149 )
eg 5 3 35 0y x 5 3 35y x 5 5 5
( all of both sides by 5)
-37
5y x
-3we can now see the gradient is and the yintercept is -7
5From this we can now find the angle the line makes with the axisx
3tan =-
5
1 3tan -
5
30.96 31
31 180 149
2626
The other form of the straight line is
ax by c o Everything on one side =0 x is positive no fractions
Rules to convert 0y mx c to ax by c get rid of fractions - whole numbers over 1
- bottoms the same- multiply both sides by the bottom
put x on the side where it is positive everything else on that side as well, other side =0
2727
Eg3
4 5
xy
20 5 4 3
20 5 4 4 5
xy
20 5 12y x
20 5 12y x
5 20 12x y
5 20 12x y
20 20 20
2828
End of Lesson 4End of Lesson 4
Ex 2.5: 1k-p, 2-5Ex 2.5: 1k-p, 2-5
Ex 2.6: allEx 2.6: all
2929
Lesson 5 & 6Lesson 5 & 6
Finding the equation of lines from given
information
3030
A (2,3), B(-1,4) and C(1,-3) form the vertices of a triangle.
Find the 1. lengths, 2. midpoints and3. gradients of all three lines (sides) in the triangle.
Revision Exercise
1. a) 3.16 b) 7.28 c) 6.08
2. a) (0.5 , 3.5) b) (0 , 0.5) c) (1.5 , 0)
3. a) -0.33 b) -3.5 c) 6
3131
y
x
2
2
4
4
6
6
– 2
– 2
– 4
– 4
– 6
– 6
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
We can find the equation given 3 combinations of information
3 1or y x
1)If we know the gradient (m) and one point (x1,y1)
Finding the equation of lines from given information
1
1
( )
( )
y ym
x x
1 1( ) ( )x x m y y
1 1( ) ( )y y m x x
3, point(2,5)m1 1( ) ( )y y m x x
( 5) 3( 2)y x 5 3 6y x
3 1 0x y
y
x
2
2
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
2
2
4
4
6
6
8
8
10
10
– 2
– 2
1 1( , )x y
2 2( , )x y
3232
eg
2, point(-3,4)
5m
1 1( ) ( )y y m x x
2Find the equation of the line with a gradient of that
5passes through ( 3,4)
2( 4) ( ( 3))
5y x
24 ( 3)
5y x
2 64
5 5y x
2 14
5 5y x
5 2 14
5 5 5or y x
5 2 14y x
2 5 14 0x y
3333
2 3 5 0x y
2) If you are given two points then you can find the gradient first, then do as 1 aboveeg
We would get the same answer if we used either point
Find the equation of the line that passes through (4,1) and (7,3)1x 1y 2x 2y
3 1
7 4m
2 1
2 1
y ym
x x
2
3m
eg(7,3)
1( ) ( )y y m x x 2
( 3) ( 7)3
y x 2 14
33 3
y x
2 5
3 3y x 3 2 5
3 3 3or y x
1 1or (4,1) ( ) ( )y y m x x
3 2 5y x
2( 1) ( 4)
3y x
2 81
3 3y x
2 5
3 3y x
First find the
gradient
Then using either point and
1 1( )y y m x x
3434
3) A (2,3), B(-1,4) and C(1,-3) form the vertices of a triangle.
Find the equations of all three lines in the triangle.
3535
2 2 0x y
4) If we are given the angle with the x axis and a point. This is similar to being given the gradient (Extn)
Summary: We can find the equation of a straight line given
1) gradient (m) and one point 2) two points, then you can find the gradient first, then do as 1 above3) the angle with the x axis and a point.
1 1( , )x y
Eg A line passes through (3,4) and makes an angle of 63.4 with the
positive direction of the axisx
tan 63.4 m2m
1 1( ) ( )y y m x x
4 2( 3)y x 4 2 6y x
2 2y x
3636
End of Lesson 5 & 6End of Lesson 5 & 6
Ex 2.7: allEx 2.7: allWB P178 #36-51WB P178 #36-51
Ex 2.8: 1h,I,j, 2all, 3-6Ex 2.8: 1h,I,j, 2all, 3-6Ex 2.9: allEx 2.9: all
3737
Lesson 7Lesson 7
Parallel Lines
3838
y mx c
Parallel Lines Are always the same distance apartNever meet
If a line is written in the form
We can see
If lines have the same gradient then they are parallel
2 6 is parallel to 2 1y x y x
Any line with a gradient of 2 will be parallel to these
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
Parallel Lines
-interceptyGradient
y
x
3939
1 2 1 2If lines two lines and are parallel then y m x c y m x c m m
3 4 5 0y x Eg Find the line through (5,7) parallel to
3 4 41 0y x
First make the subject so line is in the form of
. Then we can see the gradient
y
y mx c
3 4 5 0y x
3 4 5y x 4 5
3 3y x
4
3m
then using point(5,7)
1 1
and
( ) ( )y y m x x
4( 7) ( 5)
3y x
4 207
3 3y x
4 41
3 3y x
3 4 41
3 3 3or y x
3 4 41y x
4040
If a line through (3,4) and (6,a) is parallel to 4 2 5 0
Find "a"
x y
If lines are parallel then they must have the same gradient
4 2 5 0x y
2 4 5y x 5
22
y x
2m
2 1
2 1
( ),
( )
y ym
x x
( 4)2
(6 3)
a
The gradient of the two
points will be the same
42
3
a
6 4a 10a
4141
End of Lesson 7End of Lesson 7
Ex 2.10: 2-7Ex 2.10: 2-7
4242
Lesson 8Lesson 8
Perpendicular Lines
4343
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
y
x
y
x
2 4y x
Perpendicular means: at right angles toIf lines are perpendicular then they are at right angles to each other
-ve +veWe know
2 1y x
Perpendicular Lines
We can see these lines are not perpendicular
4444
1 22 1
1 1ie andm m
m m
Consider gradients of
12 and -
2m m
y
x
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2
2
4
4
6
6
8
8
10
10
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
2 6y x
13
2y x
The gradients are negative reciprocals of each other
1 2If two lines and are perpendicular
then
y m x c y m x c
1 2 1 m m
y
x
y
x
We can see these two
lines are perpendicular
4545
Egs Ex 2.11
Find the gradient perpendicular to 3
-1 gradient is
3
7Find the gradient perpendicular to
2
-2 gradient is
7
Find the gradient perpendicular to p
-1 gradient is
p
4646
A (2,3), B(-1,4) and C(1,-3) form the vertices of a triangle.
Find the equations of the perpendicular bisectors of all three lines
4747
Egs Ex 2.11
Eg Find equation of a line through (-2,3) perpendicular to 3 4 7 0y x 3 4 7y x
4 7
3 3y x 4
3m
3gradient
4
1
equation of perpendicular line is
( ) ( )y y m x x
3( 3) ( ( 2))
4y x
33 ( 2)
4y x
3 63
4 4y x
3 3
4 2y x
4 3 3 2
4 4 2 2or y x
4 3 6y x
3 4 6 0x y
4848
End of Lesson 8End of Lesson 8
Ex 2.11:#1-3,5-10 Extn 11-13Ex 2.11:#1-3,5-10 Extn 11-13
WB p182 # 52-59WB p182 # 52-59
4949
Lesson 9Lesson 9
Collinear points and Drawing lines
5050
Collinear PointsIe points on the same lineIf points are on the same line, any two should give the same gradient
y
x
2
2
4
4
6
6
– 2
– 2
– 4
– 4
– 6
– 6
2
2
4
4
6
6
– 2
– 2
– 4
– 4
– 6
– 6
– 8
– 8
– 10
– 10
– 12
– 12
A
B
C
gradient of AB = gradient of AC = gradient of BCSo these points are collinear
Collinear points and Drawing lines
(-3,-11)
(0,-2)
(2,4)
1
1
using y y
mx x
( 2 ( 11))
(0 ( 3))ABm
9
3
3
( 4 ( 11))
(2 ( 3))ACm
( 4 ( 2))
(2 (0))BCm
15
5
3
6
2
3
5151
NOTE: if points are collinear ie on the same line we can solve for an unknown
(2,1),( 3, ),(1, 2)are colinear so all points must meet the same equation
and have the same gradient
q
2 1
2 1
( ),
( )
y ym
x x
1 1( ) ( )y y m x x
The gradient of any two point will be the same
( 2 1)
(1 2)m
( 2 ( ))
(1 ( 3))
qm
3
1m
3m
2
1 3
qm
2
4
qm
23
4
q
2 12q
14q
14q
As points are collinear these gradients must be equal
5252
Drawing lines:To draw lines we need two pieces of information1) 2 points- plot and join2) 1 point and the gradient-plot point and use gradient to find others3) equation – either sub in x=1,2,3 etc -or use gradient intercept - or use intercept-intercept method
For Ex 2.13 we will rearrange in the form of y=mx+c to establish the gradient and y intercept
5353
End of Lesson 9End of Lesson 9Ex2.12: allEx2.12: all
Drawing LinesDrawing Lines: : Ex 2.13 oddEx 2.13 odd
WB P186 #64-75WB P186 #64-75WB P187 #76-81WB P187 #76-81
5454
Lesson 10Lesson 10
Intercept – intercept method
5555
To find x intercept, put y = 0 Remember equation of x axis is y = 0To find y intercept, put x = 0 equation of y axis is x = 0
Eg.
Intercept – intercept methodIf given the line in the form it is easier to plot using thi0 s methodax by c
2 3 6x y 1)Cuts axis put 0y x
2(0) 3 6y 3 6y
2y This gives us (0,2)
y
x
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
1
1
2
2
3
3
4
4
– 1
– 1
– 2
– 2
2)Cuts axis put 0x y
2 3(0) 6x
2 6x 3x
This gives us (3,0)
5656
22
3y x
If given the graphWe can read the equation of the graph
1) gradient intercept method
2) Find gradient
y
x
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
1
1
2
2
3
3
4
4
– 1
– 1
– 2
– 2
2 1
2 1
y ym
x x
(0 2)
(3 0)m
2
3m
Use gradient and point (3,0)
1( ) ( )y y m x x 2
( 0) ( 3)3
y x
22
3y x
3 2 32
3 3 3or y x
3 2 6y x
2 3 6 0x y
5757
End of Lesson 10End of Lesson 10
Ex 2.14allEx 2.14all
Ex 2.15 allEx 2.15 all
5858
Lesson 11Lesson 11
Simultaneous Equations
5959
Simultaneous equations represent 2 lines on a graph that may intersect
We have three methods of finding the point of intersection
1) Graphing- either plot or use calculator2) Elimination3) Substitution (covered tomorrow)
Simultaneous Equations
6060
2)Elimination
Lines intersect at (0, 1)
Line up equations x over x y over y
Sometimes we may need to rearrange to do thisDecide what to eliminate.
i.e. what number is, or can we get, the same
Do we need to add or subtract to eliminate?7 1x y 4 3 3x y
4 3 3x y 7 1x y 1
2
First write out equations across the
page
3 (7 1)
4 3 3
x y
x y
Multiply equation 1 by 3 to make the y term the
same21 3 3
4 3 3
x y
x y
25 0x 0x
Now the y terms are the same but have
opposite signs so we can add
7(0) 1y 1y
1y Sub y=-1 back into
the second equation to
check
4(0) 3( 1) 3 Correct
Sub x=0 back into
equation 1
6161
Examples
4 3 25
2 5 7
y x
y x
*
35 and 2 6
4y x y x
On Graphics Calculator either;• Graph equations then - G-solve, intersect, Answer (4,2)• Solve Simultaneous Equations in Equation Mode
6262
4 7 20
10 5 25
x y
x y
*
6363
Ex 2.18 #2Ex 2.18 #2
4 2 38
5 21 2
42 3
x y
x y
6464
End of Lesson 11End of Lesson 11
Ex 2.17:1-17 oddEx 2.17:1-17 odd
Ex 2.18 1-5 Extn allEx 2.18 1-5 Extn all
6565
Lesson 12 & 13Lesson 12 & 13
Simultaneous Equations
- Substitution
6666
3) SubstitutionIf we know what x or y is, then in the second equation, we can put what x or y is instead of writing x or y. Hence we have the whole equation with only one variable.
Eg.
Simultaneous Equations
5 4 ,y x 10 7 1 x y
5 4 ,y x 10 7 1 x y
First write out equations across
the page. It is sensible to write the
y= equation first
12
Sub 1 into 2
10 7(5 4 ) 1x x
10 35 28 1x x
18 36x
2x Now we sub x=2 back into the first equation
5 4(2)y 3y
Now we can sub y=-3 into the second equation to
check
10(2) 7( 3) 1 20 21 1
Correct
Point of intersection (2,-3)
6767
4
3 2 2 0
y x
x y
*
ExamplExamplee
6868
ExampleExample
3 9
4 13
y x
x y
*
6969
ExampleExample
2 3
9
y x
y x
*
7070
ExampleExample
3 1
3 5
y x
y x
*
Lines have the same gradient
They are parallel.
No solution possible
7171
End of Lesson 12 & 13End of Lesson 12 & 13
Ex 2.19 #1-15 odd Extn 16-18Ex 2.19 #1-15 odd Extn 16-18
Applications: Extra sheet Applications: Extra sheet
Ex 5.5 1-14 Extn allEx 5.5 1-14 Extn all
7272
Lesson 14 & 15Lesson 14 & 15
Applications of Simultaneous
Equations
7373
End of Lesson 14 & 15End of Lesson 14 & 15Applications:Applications:
Ex 2.20: 1-15 Extn allEx 2.20: 1-15 Extn allEx 16.17 sheetEx 16.17 sheet
RUR QRUR QMerit & Excellence QuestionsMerit & Excellence Questions
WB P 192 #1-7WB P 192 #1-7RUR QRUR Q
Exam Excellence Q 2.4 W/SExam Excellence Q 2.4 W/S
7474
StartersStarters
Lesson 1Lesson 1 Lesson 2Lesson 2 Lesson 3Lesson 3 Lesson 4Lesson 4 Lesson 5Lesson 5 Lesson 6Lesson 6 Lesson 7Lesson 7 Lesson 8Lesson 8
Lesson 9Lesson 9 Lesson 10Lesson 10 Lesson 11Lesson 11 Lesson 12Lesson 12 Lesson 13Lesson 13 Lesson 14Lesson 14 Lesson 15Lesson 15
7575
Starter Lesson 1
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
Find the equation and list features
7676
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1( 1)( 3)( 5)
3y x x x
( 1)( 3)( 5)
(0, 5)
5 ( 1)(3)(5)
5 15
1
3
y k x x x
Sub in
k
k
k
(0,-5)
Answers Starter Lesson 1
7777
Starter Lesson 2y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Find the equation and list features
3) Sketch the graph of 2 1, 3y x x
2) Find the midpoint of A(-2,-5) B(3,1)
and the distance between the two points
7878
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
2( 2)( 1)( 5)
5y x x x
( 2)( 1)( 5)
(0, 4)
4 ( 2)(1)(5)
4 10
2
5
y k x x x
Sub in
k
k
k
(0,-4)
1)Answers Lesson 2
7979
1 2 1 2
A ( 2, 5), B (3,1)
,2 2
2 3 5 1,
2 2
1 4,
2 2
1, 2
2
x x y ymp
2 22 1 2 1
2 2
2 2
2 2
( ) ( )
((3) ( 2)) ((1) ( 5))
(3 2) (1 5)
(5) (6)
25 36
61
7.8(1 )
d x x y y
dp
2)
Answers Lesson 2Answers Lesson 2
8080
y
x
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
We have a hole at x=32 1, 3y x x
Answers Lesson 2Answers Lesson 2
8181
1) Accurately sketch the graph
Starter Lesson 3
2 22) What transformation maps 9 onto 9y x y x
3) Sketch the graph of 3 5, 2y x x
2 1
5
xy
x
That means properly
1) Cuts axis 0
2) Cuts axis 0
3)Vertical asymptote
4) Horizontal asymptote
x y
y x
y
x
8282
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
– 11
– 11
– 12
– 12
– 13
– 13
– 14
– 14
– 15
– 15
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Cuts axis 0
2 1 0
2 1
1
22) Cuts axis 0
2(0) 1
(0) 5
1
5
x y
x
x
x
y x
y
y
1
5
4) Horizontal asymptote
22 1
5
2 2
12 2
xx x
xx x
x
y y
y y
y y
3)Vertical asymptote
5 0
5
y
x
x
2 1
5
xy
x
Q1)
8383
2) y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
2 9y x
29y x
Reflection in the x axis
8484
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
– 11
– 11
– 12
– 12
– 13
– 13
– 14
– 14
– 15
– 15
We have a hole at x = -23 5, 2y x x
8585
1) Accurately sketch the graph
Starter Lesson 4
2 22) What transformation maps ( 2) onto ( 2)y x y x
33) Sketch the graph of 4, 1
4y x x
3
4
xy
x
That means properly
1) Cuts axis 0
2) Cuts axis 0
3)Vertical asymptote
4) Horizontal asymptote
x y
y x
y
x
8686
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
– 11
– 11
– 12
– 12
– 13
– 13
– 14
– 14
– 15
– 15
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Cuts axis 0
3 0
3
2) Cuts axis 0
(0) 3
(0) 4
3
4
x y
x
x
y x
y
y
3
4
4) Horizontal asymptote
3
4
1
11 1
xx x
xx x
x
y y
y y
y y
3)Vertical asymptote
4 0
4
y
x
x
3
4
xy
x
Q1)
8787
2) y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
2( 2)y x
2( 2)y x
Reflection in the x axis
8888
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
We have a hole at x = 13
4, 14
y x x
8989
1) Accurately sketch the graph
2) Sketch the graph of ( 1)(3 )( 2)y x x x
2 1
4
xy
x
That means properly
1) Cuts axis 0
2) Cuts axis 0
3)Vertical asymptote
4) Horizontal asymptote
x y
y x
y
x
List Features
Starter Lesson 5
9090
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
– 11
– 11
– 12
– 12
– 13
– 13
– 14
– 14
– 15
– 15
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Cuts axis 0
2 1 0
2 1
1
22) Cuts axis 0
2(0) 1
(0) 4
1
4
x y
x
x
x
y x
y
y
2 1
4
4) Horizontal asymptote
2 1
4
2 2
12 2
xx x
xx x
x
y y
y y
y y
3)Vertical asymptote
4 0
4
y
x
x
2 1
4
xy
x
Q1)
9191
Features
1 intercept
21
intercept 4
Vertical asymptote 4
Horizontal asymptote 2
fundemental discontinuity at -4
point symmetry at (-4,2)
axis of symmetry 6, 2
, 2
, 2
x x
y y
x
y
x
y x y x
x y
x y
9292
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
( 1)(3 )( 2)y x x x
9393
1) Accurately sketch the graph
2 22) Sketch the graph of ( 1) 25x y
List Features
3 4
2
xy
x
That means properly
1) Cuts axis 0
2) Cuts axis 0
3)Vertical asymptote
4) Horizontal asymptote
x y
y x
y
x
List Features
Starter Lesson 6
9494
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Cuts axis 0
3 4 0
3 4
4
32) Cuts axis 0
3(0) 4
(0) 2
2
x y
x
x
x
y x
y
y
3 4
2
4) Horizontal asymptote
3 4
2
3 3
13 3
xx x
xx x
x
y y
y y
y y
3)Vertical asymptote
2 0
2
y
x
x
3 4
2
xy
x
Q1)
9595
Features
4 intercept
3 intercept 2
Vertical asymptote 2
Horizontal asymptote 3
fundemental discontinuity at -2
point symmetry at (-2,3)
axis of symmetry 5, 1
, 3
, 3
x x
y y
x
y
x
y x y x
x y
x y
9696
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
2 2( 1) 25x y
Centre(0,1)
Radius =5units
Y intercepts (0,6)and(0,-4)
Max(0,6), Min(0,-4)
2 2 025 translated by
1x y
9797
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
Q1)Find the equation, and list features
Starter Lesson 7Starter Lesson 7
9898
Features
intercept 4.6
2 intercept 7
3Vertical asymptote 3
Horizontal asymptote 5
fundemental discontinuity at 3
point symmetry at (3,5)
axis of symmetry 2, 8
, 3
, 3
x x
y y
x
y
x
y x y x
x y
x y
9999
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
2 2 2 2
2)What transformation will map
1 on to the graph 16x y x y
100100
Image radius
Object radius
4
14
Enlargement about (0,0) of scale factor 4
SF
SF
SF
Must have
centre
101101
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
2 23) What transformation maps on to ( 3) 5y x y x
Translation by
3the vector
-5
102102
1) Prove that A(-4,-4), B(1,6), C(11,1) form a right angled isosceles triangle
Starter Lesson 8Starter Lesson 8
103103
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
(-4.-4)
(1,6)
(11,1)
A
BC
1) Prove that A(-4,-4), B(1,6), C(11,1) form a right angled isosceles triangle
104104
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
(-4.-4)
(1,6)
(11,1)
1
1
( 4, 4), (1,6), (11,1)
6 ( 4)
1 ( 4)
10
52
1 6
11 15
101
2
AB
BC
y yA B C m
x x
m
m
A
B
C
12 1 as gradients multiply to give -1
2they are at right angles
105105
2 21 1
2 2
2 2
( ) ( )
( 4 1) ( 4 6)
( 5) ( 10)
25 100
125
AB
AB
AB
AB
AB
d x x y y
d
d
d
d
2 21 1
2 2
2 2
( ) ( )
(1 11) (6 1)
( 10) (5)
100 25
125
BC
BC
BC
BC
BC
d x x y y
d
d
d
d
2 21 1
2 2
2 2
( ) ( )
( 4 11) ( 4 1)
( 15) ( 5)
225 25
250
AC
AC
AC
AC
AC
d x x y y
d
d
d
d
Because 2 sides are equal and one is not equal, therefore the triangle is isosceles
We proved before that AB is Perpendicular to BC so this is a right angled isosceles triangle
106106
1)Find the equation of the line through
(9,1) and (4,7)
2)Find the equation of the line through
(3,2) and (3,10)
Starter Lesson 9Starter Lesson 9
107107
1 1
Equation
( )
61 ( 9)
56 54
15 5
6 59
5 55 6 59 0
y y m x x
y x
y x
y x
y x
2 1
2 1
1)Find the equation of the line through
(9,1) and (4,7)
7 1
4 96
5
y ym
x x
m
m
108108
2 1
2 1
2)Find the equation of the line through
(3,2) and (3,10)
10 2
3 3
We know lines with an undefined gradient
arevertical and so
In this case 3
y ym
x x
m
m undefined
x
x
109109
1) Find the line through the point (5,4) parallel
to the line 3 2 0x y
32)Draw the lines 5
44
23
3)Prove the above lines are perpendicular
y x
y x
Starter Lesson 10Starter Lesson 10
110110
1) Find the line through the point (5,4) parallel
to the line 3 2 0x y
1 1
3 2 3 (5,4)
( )
4 3( 5)
4 3 15
3 11
y x gdt pt
y y m x x
y x
y x
y x
111111
32)Draw the lines 5
44
23
3)Prove the above lines are perpendicular
y x
y x
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
3 41
4 3As gradients multiply to = -1
Lines are perpendicular
112112
Starter Lesson 11Starter Lesson 11
1) Find the equation of the line that passes
through (1,8) and (6,8)
2) Use the intercept- intercept method to plot
6 4 3x y
3) Find the line through the point (1,4)
perpendicular to the line 4 3 0x y
113113
2 1
2 1
1) Find the equation of the line that passes
through (1,8) and (6,8)
8 8
6 10
50
y ym
x x
m
m
m
Line has a gradient of zero so horizontal
?
both y coordinates are 8
8
y
y
114114
2) Use the intercept- intercept method to plot
6 4 3
Cuts axis 0
6 4(0) 3
6 3
3
61
2
x y
x y
x
x
x
x
Cuts axis 0
6(0) 4 3
4 3
3
4
y x
y
y
y
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
115115
3) Find the line through the point (1,4) perpendicular
to the line 4 3 0
4 3
4 3
4
1
4
x y
x y
y x
m
m
1 1
1(1,4)
4( )
14 ( 1)
41 1
44 4
1 1 16
4 4 41 17
4 44 17 0
m
y y m x x
y x
y x
y x
y x
x y
116116
Lesson 12 StarterLesson 12 Starter1) Solve
2 2
5 2 10 0
y x
x y
3 12) Sketch
2and list features
xy
x
That means properly
1) Cuts axis 0
2) Cuts axis 0
3)Vertical asymptote
4) Horizontal asymptote
x y
y x
y
x
117117
1) Solve
2 2 5 2 10 0 2 2
5 2 10 0 5(3) 2 10 0 2(2.5) 3 2
15 2 10 0 5 3 2
2 2 5 2
5 2 10 2.5
4 12
3
y x x y y x
x y y
y
x y y
x y y
x
x
true
(3,2.5)
Checked on the
calculator of course
118118
3 12) Sketch and list features
2
xy
x
y
x
1
1
2
2
3
3
4
4
5
5
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
– 11
– 11
– 12
– 12
– 13
– 13
– 14
– 14
– 15
– 15
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
– 9
– 9
– 10
– 10
1) Cuts axis 0
3 1 0
3 1
1
32) Cuts axis 0
3(0) 1
(0) 2
1
2
x y
x
x
x
y x
y
y
3 1
2
4) Horizontal asymptote
3 1
2
3 3
13 3
xx x
xx x
x
y y
y y
y y
3)Vertical asymptote
2 0
2
y
x
x
3 1
2
xy
x
119119
Features
1 intercept
31
intercept 2
Vertical asymptote 2
Horizontal asymptote 3
fundamental discontinuity at 2
point symmetry at (-2,3)
axis of symmetry 5, 1
, 3
, 3
x x
y y
x
y
x
y x y x
x y
x y
120120
Starter Lesson 13Starter Lesson 13
1) Find the line perpendicular to 5 4 7 0 which
passes through the point (3,1)
2) If (3,4) (6,2) (8,a) are collinear
Find a
y x
121121
1 1
1) Find the line perpendicular to 5 4 7 0 which
passes through the point (3,1)
5 -4 7
4 7
5 54 5
,5 4
( )
51 ( 3)
45 11
4 45 4 11 0
y x
y x
y x
m m
y y m x x
y x
y x
x y
122122
2 1
2 1
2) If (3,4) (6,2) (8,a) are collinear
Find a
if lines are collinear they have the same gdt
2 4 2 2
6 3 3 8 62 2 2 2 3
3 2 3 2 2
4 3( 2)
4 3 6
2 3
2
3
y ym
x x
am
am
a
a
a
a
123123
Starter Lesson 14Starter Lesson 14
1)Show that the line through (0,4) and (4,1) is
perpendicular to the line through (3,6) and (-3,-2)
2)Find the midpoint of A(4,2) and B(10,10)
124124
2 1 2 1
2 1 2 1
1)Show that the line through A(0,4) and B(4,1) is
perpendicular to the line through C(3,6) and D(-3,-2)
1 4 2 6
4 0 3 33 8
4 64
31
3 41 line
4 3
AB CD
AB CD
AB CD
CD
AB CD
y y y ym m
x x x x
m m
m m
m
as m m
s are perpendicular
125125
1 2 1 2
2)Find the midpoint of A(4,2) and B(10,10)
( ) ( ) ,
2 2
(4 10) (2 10),
2 2
14 12,
2 2
7,6
x x y ym
m
m
m
126126
Starter Lesson 15Starter Lesson 15
Prove that the points A(-3,7) B(9,15) C(5,-5)
form a right angled isosceles triangle
127127
y
x
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
12
13
13
14
14
15
15
16
16
17
17
18
18
– 1
– 1
– 2
– 2
– 3
– 3
– 4
– 4
– 5
– 5
– 6
– 6
– 7
– 7
– 8
– 8
Prove that the points A(-3,7) B(9,15) C(5,-5)
form a right angled isosceles triangle
B
A
C
2 22 1 2 1
2 2
2 2
2 22 1 2 1
2 2
2 2
2 22 1 2 1
2 2
2 2
( ) ( )
(9 ( 3)) (15 7)
12 8
208
( ) ( )
(5 ( 3)) ( 5 7)
8 12
208
( ) ( )
(9 5) (15 ( 5)
4 20
416
AB
AC
BC
d x x y y
d
d
d
d x x y y
d
d
d
d x x y y
d
d
d
As two sides are the same and one is different,
this proves that the triangle is isosceles
128128
2 1
2 1
15 7
9 ( 3)
8
122
3
AB
y ym
x x
2 1
2 1
5 7
5 ( 3)
12
83
2
AC
y ym
x x
2 3
3 21
Triangle is a right angled isosceles
AB ACm m
ABis toAC