co-ordinate geometry objectives: students should be able to * find the distance between two points *...

Co-ordinate geometry

Objectives: Students should be able to

* find the distance between two points

* find the gradient of a line

* find the mid-point of a line

Co-ordinate geometry Co-ordinates

Co-ordinates are a means of describing a position relative to some fixed points, or origin. In two dimensions you need two pieces of information; in three dimensions, you need three pieces of information.

•

• x

yA

B

The coordinate of A ? (4, 3)

The coordinate of B ? ( -4, -1)

The mid – point of a line segment

1 2 1 2

2 2

x x y y( )M ,

-6

-4

-2

0

2

4

6

8

10

-3 -2 -1 0 1 2 3 4 5

Example 1: Find the midpoint of PQ where P is the point (2, 4) and Q is the point (4, 8).

2 4 4 8

2 2,

Midpoint = = (1, 2)

The coordinates of the midpoint are (1, 2)

The mid – point of a line segment

Example 2. Find a and b if the point (3, 5) is the midpoint of the line joining (3a, 2a 4b) and (a, 3a + 2b).

33 4 6 1 5

2

a aa a .

2 4 3 25

2

a b a b 5 2 10a b

2b = 10 5a 2b = 10 7.5 = 2.5 b = 1.25

1 2 1 2

2 2

x x y y( ),( )M

x coordinate:

y coordinate:

Therefore a = 1.5 and b = 1.25

The gradient of a line The direction of a straight line is given by its gradient.

The gradient of a line is the amount by which the y coordinate increases if we move along the line far enough to increase the x coordinate by one unit.

i.e. the gradient of a line is a measure of its steepness. The steeper the line, the larger the gradient.

Example

P

Q

Negative gradientOn the line MN, as we move from A to A1, the x coordinate increases by 1 unit but the y coordinate decreases by g units. But a decrease of g units may be regarded as an increase of (– g) units. Thus the gradient of MN will be –g.

The gradient of a line joining ( x1 , x2 ) to ( y1 , y2 )

y2 – y1

x2 – x1

2 1

2 1

y y

The gradientx x

Example

If A and B are the points (1, 3) and (2, 1) find the gradient of AB.

increase in y 1 3gradient of AB 2

increase in x 2 1

Example

If P,Q, R and S are points (2, 3), (4, 8), (-3, -2) and (1, 8), respectively, show that PQ is parallel to RS.

5.224

38 PQ ofgradient

5.24

10

)3(1

)2(8 RS ofgradient

Therefore PQ is parallel to RS

The distance between two points

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6

A(1, 3)

B(5, 6)

2 22 1 2 1( x x ) ( y y ) AB =

N(5, 3)

Find the distance AB

AN = 4

NB = 3

Pythagoras:2 2 2AB AN NB 2 2 24 3AB

AB2 = 25AB = 5

The distance AB between two points (x1, y1) and B(x2, y2) is given by the formula

Example: For the pair of points A(–2, 6), B(4, –2), calculate

(a) The midpoint of AB (b) The gradient of AB

(c) The distance ABA(-2, 6)

B(4, -2)

x

y

(a) Mid-point = 1 2 1 2

2 2

x x y y( , )

=(1, 2)2 4 6 2

2 2( , )

(b) Gradient = 2 1

2 1

y y

x x

13

2 6 81

4 2 6

2 22 1 2 1( x x ) ( y y ) AB = (c) 2 24 2 2 6 36 64( ) ( ) =10

Find the vertices of the triangle by solving simultaneously the equations for each pair of lines

ExampleA triangle is formed by three straight lines, y = , 2x + y + 5 = 0and x + 3y – 5 = 0. Prove that the triangle is isosceles.

12 x

12y x

2 5 0x y (1)(2)

Let the point of intersection of line (1) and line (2) be P.

Substitute y from (1) into (2).12At 2 5 0P x x

122 5x

2x Substitute in (1) 1

2 ( 2) 1y P is the point (–2, –1).

Let the point of intersection of line (1) and line (3) be Q.Substitute y from (1) into (3).At Q, 1

23 5 0x x 122 5x

2x Substitute in (1)

12 (2) 1y

Q is the point (2, 1).

Let the point of intersection of line (2) and line (3) be R.

3 5 0x y (3)

Substitute in (2)8 5 0y

3y R is the point (–4, 3).

xy 21 (1)

052 yx (2)053 yx (3)

(2) × 3 (4)01536 yx(4) – (3) 0205 x

4x

Draw a sketch, labelling the points

Work out the squares of the lengths and compare2 2 2( 4 ( 2)) (3 ( 1))RP 2 2( 2) 4 202 2 2(2 ( 2)) (1 ( 1))PQ 2 24 2 20

2 2 Triangle is isosceles.RP PQ RP PQ RPQ

Tip: Since we can see that two lengths are the same, there is no need to work out the third length.

( 4,3)R

( 2, 1)P

(2,1)Q

y

xO

Draw a sketch

Work out AC2 and BC2

Example

In the triangle ABC, A, B and C are the points (–4, 1), (–2, –3) and (3, 2),respectively.

a) Show that ABC is isosceles.

Tip: Use the sketch to identify which two sides are likely to be the same length.

2 2 2(2 1) (3 ( 4))AC 2 21 7 50 2 2 2(2 ( 3)) (3 ( 2))BC 2 25 5 50 2 2 Triangle is isosceles.AC BC AC BC ABC

y

x

A(–4, 1)C(3, 2)

B(–2, –3)

O

Identify the base and substitute into the mid-point formula

© Pearson Education Ltd. 2005

b) Find the coordinates of the midpoint of the base.

The base of the triangle is AB.

1 2 1 2,2 2

x x y yM

4 ( 2) 1 ( 3),

2 2

)1,3(

Let the mid-point of AB be M.

y

x

A(–4, 1)C(3, 2)

B(–2, –3)

O

Show M on the diagram

c) Find the area of ABC.

Tip: Since triangle ABC is isosceles, MC is perpendicular to AB, by symmetry.

Find AB and MC2 2( 2 ( 4)) ( 3 1)AB 2 22 ( 4) 20 2 5

2 2(3 ( 3)) (2 ( 1))MC 2 26 3 45 3 5

Use the formula for the area of a triangle

Area of triangle ABC 12 AB MC 5 3 5

215 units

y

x

A(–4, 1)C(3, 2)

B(–2, –3)

O

M(–3, –1)

The properties of lines

Parallel lines have equal gradients.

Lines parallel to the x-axis have gradient zero. Lines parallel to the y-axis have infinite gradient.

Two lines are perpendicular if the product of their gradient is –1.