L2 Coordinate Geometry M.Guttormson Co Ordinate Geometry Midpoint of a Line Segment Distance...

L2 Coordinate GeometryM.Guttormson

Co Ordinate Geometry

• Midpoint of a Line Segment

• Distance• Gradients• Parallel lines• Perpendicular lines• Equation of a line

– Two points– Point and gradient

• Intersection of two straight lines

• Terms for excellence

• Rural Neighbours• Goat fences• Game park (2006 SINCOS)• Game Park (2006 SINCOS)

Answers• Exam 2006 Questions• Exam 2006 answers


Midpoint of a Line Segment

€

y = 2x+1

A

B

3

6

A

B

1.5

3

MP

€

(21

2,6)


Midpoint of a Line Segment

A line segment is a part of a line between two points. The midpoint is halfway between the points. In general, if the two points are

€

x1,y1( ) and

€

x2,y2( ) then the coordinates of the midpoint are:

€

x1+x2( )2 , y1+y2( )

2 ⎡ ⎣ ⎢ ⎤

⎦ ⎥ Eg: Find the midpoint of the line segment AB where A = (1,3) and B= (4,9)

=

€

1+42 ,3+9

2 ⎡ ⎣ ⎢ ⎤

⎦ ⎥ = (2.5, 6)


Distance

€

y = 2x+1

A

B

3

6

A

B

3

66.708


DistanceIf we are given two points on a graph it is possible to find the distance between them using Pythagoras.

€

d2=x2−x1( )2+y2−y1( )2

€

d= x2−x1( )2+y2−y1( )2

Eg: Calculate the distance between the points (1, 3) and (4, 9)

€

d2=4−1( )2+9−3( )2

=32 + 62 =9 + 36 =45

€

d= 45

€

d=6.708


Gradients

€

y = 2x+1

A

B

€

y = 2x+1

A

B

3

6

€

m = 2


Gradients

The gradient of a line is a number, which describes how steep the line is. It describes the change in y for every step in x.

€

m=y2−y1x2−x1

Eg: Find the gradient of the line joining (1, 3) and (4, 9).

€

m=9−34−1

€

m=63

m=2


Parallel lines

Parallel lines have the same direction. This means they never meet. They are like railway tracks. The gradients of parallel lines are the same. Therefore if

€

m1=m2 then the two lines they represent must be parallel.

Line A has a gradient of 2 Line B is parallel to A and goes through (5, 4) What is the equation of line B

€

y−4=2(x−5)y−4=2x−10y=2x−6

B


Perpendicular linesTwo lines are perpendicular if they meet at right angles. If two lines are perpendicular

€

m1×m2=−1 An easy way to think of this is to simply flip the gradient over and change the sign Eg:

€

m1 = 2 2 x

€

m2 = -1

€

m2 = -

€

12

€

y = 2x +1

y = −1

2x +1


Equation of a line (Two points)

The equation of a line can be expressed in the form where m is the gradient of the line and c is the y intercept.

Given two sets of coordinates, we can place them into a formula to extract the equation of the line they are on. This formula is .

• Use both sets of coordinates to find out the gradient• Use one set of coordinates in place of y1 and x1 along with the newly found gradient (m) to form the equation for the line.

Note that some rearranging of the formula is required to get it into the form

€

y = mx+ c

€

y− y1 = m(x− x1)

€

y− y1 = m(x− x1)

€

y = mx+ c


Equation of a line (Two points)

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A = (−1,−2)

B = (1,4)

m(AB) =4−−2

1−− 1m(AB) = 3

y − y1 = m(x − x1)

y − 4 = 3(x −1)

y − 4 = 3x − 3

y = 3x +1A

B


Equation of a line (Point and gradient)

The equation of a line can be expressed in the form

€

y=mx+c, where m is the gradient of the line and c is the y intercept. Given the gradient (m) and one set of coordinates, we can place them into a formula to extract the equation of the line they are on. This formula is

€

y−y1=m(x−x1). 1. Use the coordinates in place of y1 and x1 along with the gradient (m) to form the

equation for the line. Note that some rearranging of the formula

€

y−y1=m(x−x1) is required to get it into the form

€

y=mx+c


Equation of a line (Point and gradient)

B

€

A = (1,4)

m = 3

y − y1 = m(x − x1)

y − 4 = 3(x −1)

y − 4 = 3x − 3

y = 3x +1


Intersection of two straight lines

When two lines intersect (cross over each other), we can determine the coordinates of the intersection by solving a simultaneous equation. Example: Find the point of intersection of the lines

€

y=x+3 and

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y=3x+1.

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x+3=3x+1

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2=2x

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x=1

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y=(1)+3

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y=4 Coordinates of intersection are (1,4)


Terms for excellenceCollinear: Points that are on the same straight line.

Perpendicular bisector: The perpendicular bisector of a triangle is a straight line passing through the midpoint of a side and being perpendicular to it.

Circumcenter: Three perpendicular bisectors meet in a single point called the triangles circumcenter.

Altitude: An altitude of a triangle is a straight line through a vertex and perpendicular to the opposite side.

Orthocenter: The three altitudes of a triangle intersect in a single point, called the orthocenter of the triangle. The orthocenter lies inside the triangle.

Circumcenter


Rural Neighbours

Draw a set of x and y axes that range from -8 to 8. Each value represents 1km.

As you work through the exercise mark in the points and line segments on the number plane.

1. Farmhouse A is located at (-3, 7).Farmhouse B is located at (7,1).

a) Plot these points and join them with a straight line, this is a road between two houses.b) Calculate the length of the road between farmhouse A and B.

2. Farmhouse C is located at (-3,-6). A straight road leads from it to a shop represented by point S. The shop is located midway between house A and B.

a) Show how you could calculate the coordinate for the midpoint between house A and B?b) How far is it from the shop to the house C? (It is a road, round to 1dp)

3. What is the gradient of the line that represents road AB?

4. What is the gradient of the line that represents road CS?

5. Farmhouse D is located at (-7, -1). A straight road links house A to D. Is road AD parallel to road CS? Show valid reasons for your answer.

6. For each road create an equation in the form y= mx + c


Goat Fences The diagram below shows an island. The owner is planning to stock the island with three different breeds of goats. She wishes to keep these breeds apart, so she plans to build two fences AB and CD shown below. The coordinates are in metres. The diagram below is a model of the situation and not a scale diagram. You are encouraged to sketch models of each situation to help you isolate what is needed to solve the question. (260, 240) A C (360, 270) F Midpoint Irrigation Channel E B (350, 60) D (480, 30) Fences


Goat Fences

Questions

1. Are the fences parallel? Explain why or why not.2. The cost of fencing is $2.50 per metre. How much will it cost to build the two fences?

An irrigation channel EF is perpendicular (at right angles) to fence AB and passes through its midpoint.

3. What is the equation of the line that represents the irrigation channel4. What are the coordinates where the irrigation channel and the fence CD intersect?


Game park questionsQUESTION ONE

(a) Calculate the distance between the points (5,-1) and (3,4).

(b) Find the equation of the line joining the points (5, -1) and (3,4).

QUESTION TWO

Find the equation of the line that is perpendicular to the line and passes through the point (6,-3).

QUESTION THREE

A large animal game park is set up on a grid system with the entrance at E (0,0) with north in the direction of the positive y-axis.

There is a lion at the point L (3,6).

The watering hole is at the point W (7,8)

(a) The lion starts walking in a straight line from its position at L towards the watering hole at point W. When it is halfway between L and W, it turns at right angles to the line LW.

Find the equation of the line the lion is now walking on, i.e. find the equation of the perpendicular bisector of the line LW.

(b) A helicopter later sights the lion at point (8,1). At this point the lion is km from a zebra that is directly east of the watering hole at W.

Find the co-ordinates of the zebra.

€

y = 1− 3x

€

74


Game park Questions

QUESTION FOUR

Prove that the point A (3,5), B(7,13) and C(17,33) are collinear. Plotting points is NOT sufficient.

QUESTION FIVE

The altitudes of a triangle intersect at a point called orthocentre.

Find the coordinates of the orthocentre, O, of the triangle WXY where the vertices of the triangle are W (-3,-3), X(5,-3) and Y(1,5).

Plotting points is NOT sufficient.


Game park Answers


EXAM 2006 Questions

QUESTION ONE (a) Calculate the midpoint of the line joining the points (4,5) and (6,-1). (b) Find the equation of the line joining the points (4, 5) and (6, -1) (c) Find the equation of the line that is perpendicular to the line

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y=2x−3 and the passes through the point (6, -1)


EXAM 2006 QuestionsQUESTION TWO A mountain bike track has been marked by four points O (0, 0), A (-2, 2), B(2, 4) and C(6, 0) relative to a set of axes on a map.

(b) A fi rst aid tent is to be positioned at the intersection of the diagonals of the quadrilateral OABC. Find the coordinates of the first aid tent.

(c) Vanessa is checking out the track before the race. She decides to take a shortcut

from B back to the track OA.

If she could take the short path, fi nd the coordinates of the point where her shortcut path meets the track OA.

A(-2, 2)

O(0, 0) C(6, 0)

B(2, 4)


EXAM 2006 Questions

QUESTION THREE Prove that the triangle with vertices A (1, 1), B (3, 2) and C (0, 8) is a right-angled triangle. Plotting points is not sufficient. QUESTION FOUR A (8, 1) and B (k, -4) are two points. The perpendicular bisector of AB cuts the y-axis at -3. Find the two possible values of k.


EXAM 2006 ANSWERS

QUESTION ONE (a)

€

MP=(4+62 ,5+−1

2 )MP=(5,2) (b)

€

m=−1−56−4

m=−3

y−5=−3(x−4)y−5=−3x+12y=−3x+17

(c)

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m1=2m2=−12

y+1=−12(x−6)y+1=−12x+3y=−12x+2


EXAM 2006 ANSWERS

QUESTION TWO (a)

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AC(m)=0−26+2

AC(m)=−14

y−2=−14(x+2)

y−2=−14x−

12

y=−14x+

32

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OB(m)=0−40−2

OB(m)=2

y−0=2(x+0)y=2x

Intersect

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2x=−14x+

32

214x=32

94x=

32

x=23

y=43


EXAM 2006 ANSWERSQUESTION TWO (b)

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OA(m1)=0−20+2=−1

m2=1

y−4=1(x−2)y−4=x−2y=x+2

OAy−2=−1(x+2)y−2=−x−2y=−x

Equation of shortest root

Intersection of OA and shortest root

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−x=x+2−2=2xx=−1y=1


EXAM 2006 ANSWERS

QUESTION THREE

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d(AB)= (1−3)2+(1−2)2d(AB)= 4+1d(AB)= 5

d(AC)= (1−0)2+(1−8)2d(AC)= 1+49d(AC)= 50

d(BC)= (3−0)2+(2−8)2d(AC)= 9+36d(AC)= 45

a2+b2=c25+45=50

€

m(AB)×m(BC) = −1

2−1

3−1×

8− 2

0− 3= −1

1

2×

6

−3= −1


EXAM 2006 ANSWERS

€

m =1− (−4)

8 − k

y −1=5

8 − k(x − 8)

y =5

8 − k(x − 8) +1

8 + k

2,−

3

2

⎛

⎝ ⎜

⎞

⎠ ⎟

m1 ×m2 = −1

5

8 − k×m2 = −1

m2 =−8 + k

5

€

y +3

2=

−8 + k

5x −

8 + k

2

⎛

⎝ ⎜

⎞

⎠ ⎟

−3+3

2=

−8 + k

50 −

8 + k

2

⎛

⎝ ⎜

⎞

⎠ ⎟

−3

2= −

1

10(−8 + k)(8 + k)

15 = k 2 − 64

k 2 = 79

k = ± 79

Gradient

Gradient in line equation

Rearranged

Midpoints

Perpendicular Gradients

Using midpoints

Using (3, 0)Rearrange & solve

QUESTION FOUR

L2 Coordinate Geometry M.Guttormson Co Ordinate Geometry Midpoint of a Line Segment Distance...

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