6 mat unit 4 coordinate geometry

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

Co-ordinate Co-ordinate GeometryGeometry

Lesson 1Lesson 1Midpoint of a line and Midpoint of a line and

distance between 2 pointsdistance between 2 points

1 2 1 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

( ) ( ),

( 4 0) (0 8),

x y x y

x x y ymp

(x2,y2)

Midpoint of a lineMidpoint of a line

)( 2121 yyxx

The midpoint of the line joining two points (x1,y1) and (x2,y2) is

1 1( , )x y

Midpoint of a Line

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( ) ( ),

x x y ymp

(6 2) ( 1 3)

(6 2) ( 1 3),

4 ( 4),

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 ) (9 )4 , 5

8 2 , 10 9x y

6, 1x y

(6,1)B

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

Ex 2.1: #1a-I (1column) 2-5Ex 2.1: #1a-I (1column) 2-5WB P170#12-25 WB P170#12-25

(15mins) (15mins)

2 2 22 1 2 1

2 22 1 2 1

( ) ( )

d x x y y

(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)

2 2 22 1 2 1

2 22 1 2 1

( ) ( )

d x x y y

(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

(y2-y1)

(x2-x1)

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

= 4z units

x units

p units

Remember

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

Lesson 2Lesson 2

To find the gradient of a To find the gradient of a lineline

Gradient

(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

(0 ( 4))m

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

(( 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)( 6) 3p p 2 10 24 3p p 2 10 21 0p p

( 7)( 3) 0p p 7, 3p p

WB P174 #26-35Ex 2.3: # 2-9

Lesson 3Lesson 3

Gradient of a line making an angle with the axes

(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

A Gradient =

Atan m

1 1 2 2, ( 4,0), , 0,8x y x y 1x 1y 2x 2y

(0 ( 4))Tan

2Tan 1(2)Tan

tany y

36.9 143.1

eg find the angle a line with the gradient of – ¾ makes with the positive direction of the x axis

1 3tan

smallest angle is 36.9°

Largest angle is 36.9 180 143.1

tan (shift tan 9999999)

gradient undefined

Eg Ex 2.4

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

Ex 2.4: #1 odd, 2 all, 3 odd, 4 Ex 2.4: #1 odd, 2 all, 3 odd, 4 -7-7

Lesson 4Lesson 4

Rearranging the equation of a straight

If a line is written in the form y mx c

3we can now see the gradient is and the intercept is -1

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

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)

2 3 2 0y x

( 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)

-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 =-

1 3tan -

30.96 31

31 180 149

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

20 5 4 3

20 5 4 4 5

20 5 12y x

5 20 12x y

20 20 20

Ex 2.5: 1k-p, 2-5Ex 2.5: 1k-p, 2-5

Ex 2.6: allEx 2.6: all

Lesson 5 & 6Lesson 5 & 6

Finding the equation of lines from given

information

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

– 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( ) ( )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

1 1( , )x y

2 2( , )x y

2, point(-3,4)

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))

24 ( 3)

5 5y x

5 2 14

5 5 5or y x

5 2 14y x

2 5 14 0x y

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

eg(7,3)

1( ) ( )y y m x x 2

( 3) ( 7)3

y x 2 14

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)

3 3y x

First find the

gradient

Then using either point and

1 1( )y y m x x

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.

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

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

Lesson 7Lesson 7

Parallel Lines

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

– 10

Parallel Lines

-interceptyGradient

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 mx c

3 4 5 0y x

3 4 5y x 4 5

3 3y x

then using point(5,7)

( ) ( )y y m x x

4( 7) ( 5)

3 3y x

3 4 41

3 3 3or y x

3 4 41y x

If a line through (3,4) and (6,a) is parallel to 4 2 5 0

Find "a"

If lines are parallel then they must have the same gradient

4 2 5 0x y

2 4 5y x 5

The gradient of the two

points will be the same

6 4a 10a

Ex 2.10: 2-7Ex 2.10: 2-7

Lesson 8Lesson 8

Perpendicular Lines

– 10

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

1 22 1

1 1ie andm m

Consider gradients of

12 and -

– 10

2 6y x

The gradients are negative reciprocals of each other

1 2If two lines and are perpendicular

y m x c y m x c

1 2 1 m m

We can see these two

lines are perpendicular

Egs Ex 2.11

Find the gradient perpendicular to 3

-1 gradient is

7Find the gradient perpendicular to

-2 gradient is

Find the gradient perpendicular to p

-1 gradient is

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

Egs Ex 2.11

Eg Find equation of a line through (-2,3) perpendicular to 3 4 7 0y x 3 4 7y x

3 3y x 4

3gradient

equation of perpendicular line is

( ) ( )y y m x x

3( 3) ( ( 2))

33 ( 2)

4 4y x

4 2y x

4 3 3 2

4 4 2 2or y x

4 3 6y x

3 4 6 0x y

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

Lesson 9Lesson 9

Collinear points and Drawing lines

Collinear PointsIe points on the same lineIf points are on the same line, any two should give the same gradient

– 10

– 12

gradient of AB = gradient of AC = gradient of BCSo these points are collinear

Collinear points and Drawing lines

(-3,-11)

(0,-2)

using y y

( 2 ( 11))

(0 ( 3))ABm

( 4 ( 11))

(2 ( 3))ACm

( 4 ( 2))

(2 (0))BCm

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

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))

As points are collinear these gradients must be equal

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

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

Lesson 10Lesson 10

Intercept – intercept method

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

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)

2)Cuts axis put 0x y

2 3(0) 6x

2 6x 3x

This gives us (3,0)

If given the graphWe can read the equation of the graph

1) gradient intercept method

2) Find gradient

(3 0)m

Use gradient and point (3,0)

1( ) ( )y y m x x 2

( 0) ( 3)3

3 2 32

3 3 3or y x

3 2 6y x

2 3 6 0x y

Ex 2.14allEx 2.14all

Ex 2.15 allEx 2.15 all

Lesson 11Lesson 11

Simultaneous Equations

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)

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

First write out equations across the

3 (7 1)

Multiply equation 1 by 3 to make the y term the

same21 3 3

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

4(0) 3( 1) 3 Correct

Sub x=0 back into

equation 1

Examples

4 3 25

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

4 7 20

10 5 25

Ex 2.18 #2Ex 2.18 #2

4 2 38

5 21 2

Ex 2.17:1-17 oddEx 2.17:1-17 odd

Ex 2.18 1-5 Extn allEx 2.18 1-5 Extn all

- Substitution

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.

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

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

10(2) 7( 3) 1 20 21 1

Correct

Point of intersection (2,-3)

3 2 2 0

ExamplExamplee

ExampleExample

Lines have the same gradient

They are parallel.

No solution possible

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

Applications of Simultaneous

Equations

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

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

Starter Lesson 1

– 10

Find the equation and list features

– 10

1( 1)( 3)( 5)

3y x x x

( 1)( 3)( 5)

(0, 5)

5 ( 1)(3)(5)

y k x x x

Sub in

(0,-5)

Answers Starter Lesson 1

Starter Lesson 2y

– 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

– 10

2( 2)( 1)( 5)

5y x x x

( 2)( 1)( 5)

(0, 4)

4 ( 2)(1)(5)

y k x x x

Sub in

(0,-4)

1)Answers Lesson 2

1 2 1 2

A ( 2, 5), B (3,1)

2 3 5 1,

x x y ymp

2 22 1 2 1

( ) ( )

((3) ( 2)) ((1) ( 5))

(3 2) (1 5)

(5) (6)

7.8(1 )

d x x y y

Answers Lesson 2Answers Lesson 2

We have a hole at x=32 1, 3y x x

Answers Lesson 2Answers Lesson 2

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

That means properly

1) Cuts axis 0

2) Cuts axis 0

3)Vertical asymptote

4) Horizontal asymptote

– 10

– 11

– 12

– 13

– 14

– 15

– 10

1) Cuts axis 0

22) Cuts axis 0

2(0) 1

– 10

2 9y x

Reflection in the x axis

– 10

– 11

– 12

– 13

– 14

– 15

We have a hole at x = -23 5, 2y x x

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

That means properly

1) Cuts axis 0

2) Cuts axis 0

– 10

– 11

– 12

– 13

– 14

– 15

– 10

1) Cuts axis 0

2) Cuts axis 0

– 10

2( 2)y x

Reflection in the x axis

– 10

We have a hole at x = 13

2) Sketch the graph of ( 1)(3 )( 2)y x x x

That means properly

1) Cuts axis 0

2) Cuts axis 0

List Features

Starter Lesson 5

– 10

– 11

– 12

– 13

– 14

– 15

– 10

1) Cuts axis 0

22) Cuts axis 0

2(0) 1

Features

1 intercept

intercept 4

Vertical asymptote 4

Horizontal asymptote 2

fundemental discontinuity at -4

point symmetry at (-4,2)

axis of symmetry 6, 2

y x y x

– 10

( 1)(3 )( 2)y x x x

2 22) Sketch the graph of ( 1) 25x y

List Features

That means properly

1) Cuts axis 0

2) Cuts axis 0

List Features

Starter Lesson 6

– 10

1) Cuts axis 0

32) Cuts axis 0

3(0) 4

Features

4 intercept

3 intercept 2

fundemental discontinuity at -2

y x y x

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

Q1)Find the equation, and list features

Starter Lesson 7Starter Lesson 7

Features

intercept 4.6

2 intercept 7

3Vertical asymptote 3

fundemental discontinuity at 3

point symmetry at (3,5)

y x y x

2 2 2 2

2)What transformation will map

1 on to the graph 16x y x y

100100

Image radius

Object radius

Enlargement about (0,0) of scale factor 4

Must have

centre

101101

2 23) What transformation maps on to ( 3) 5y x y x

Translation by

3the vector

102102

1) Prove that A(-4,-4), B(1,6), C(11,1) form a right angled isosceles triangle

103103

– 10

(-4.-4)

(11,1)

1) Prove that A(-4,-4), B(1,6), C(11,1) form a right angled isosceles triangle

104104

– 10

(-4.-4)

(11,1)

( 4, 4), (1,6), (11,1)

6 ( 4)

1 ( 4)

y yA B C m

12 1 as gradients multiply to give -1

2they are at right angles

105105

2 21 1

( ) ( )

( 4 1) ( 4 6)

( 5) ( 10)

25 100

d x x y y

2 21 1

( ) ( )

(1 11) (6 1)

( 10) (5)

100 25

d x x y y

2 21 1

( ) ( )

( 4 11) ( 4 1)

( 15) ( 5)

225 25

d x x y y

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)

(3,2) and (3,10)

107107

Equation

61 ( 9)

5 55 6 59 0

y y m x x

(9,1) and (4,7)

108108

(3,2) and (3,10)

We know lines with an undefined gradient

arevertical and so

In this case 3

m undefined

109109

1) Find the line through the point (5,4) parallel

to the line 3 2 0x y

32)Draw the lines 5

3)Prove the above lines are perpendicular

110110

1) Find the line through the point (5,4) parallel

to the line 3 2 0x y

3 2 3 (5,4)

4 3( 5)

4 3 15

y x gdt pt

y y m x x

111111

32)Draw the lines 5

3)Prove the above lines are perpendicular

4 3As gradients multiply to = -1

Lines are perpendicular

112112

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

1) Find the equation of the line that passes

through (1,8) and (6,8)

Line has a gradient of zero so horizontal

both y coordinates are 8

114114

2) Use the intercept- intercept method to plot

Cuts axis 0

6 4(0) 3

Cuts axis 0

6(0) 4 3

115115

3) Find the line through the point (1,4) perpendicular

to the line 4 3 0

1(1,4)

14 ( 1)

1 1 16

4 4 41 17

4 44 17 0

y y m x x

116116

Lesson 12 StarterLesson 12 Starter1) Solve

5 2 10 0

3 12) Sketch

2and list features

That means properly

1) Cuts axis 0

2) Cuts axis 0

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

y x x y y x

(3,2.5)

Checked on the

calculator of course

118118

3 12) Sketch and list features

– 10

– 11

– 12

– 13

– 14

– 15

– 10

1) Cuts axis 0

32) Cuts axis 0

3(0) 1

119119

Features

1 intercept

intercept 2

fundamental discontinuity at 2

y x y x

120120

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

121121

1) Find the line perpendicular to 5 4 7 0 which

passes through the point (3,1)

5 -4 7

5 54 5

51 ( 3)

4 45 4 11 0

y y m x x

122122

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)

123123

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

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

3 41 line

y y y ym m

x x x x

as m m

s are perpendicular

125125

1 2 1 2

2)Find the midpoint of A(4,2) and B(10,10)

( ) ( ) ,

(4 10) (2 10),

14 12,

x x y ym

126126

Prove that the points A(-3,7) B(9,15) C(5,-5)

form a right angled isosceles triangle

127127

Prove that the points A(-3,7) B(9,15) C(5,-5)

form a right angled isosceles triangle

2 22 1 2 1

( ) ( )

(9 ( 3)) (15 7)

( ) ( )

(5 ( 3)) ( 5 7)

( ) ( )

(9 5) (15 ( 5)

d x x y y

As two sides are the same and one is different,

this proves that the triangle is isosceles

128128

9 ( 3)

5 ( 3)

Triangle is a right angled isosceles

AB ACm m

ABis toAC

6 mat unit 4 coordinate geometry

Education

Transcript of 6 mat unit 4 coordinate geometry

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