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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
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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)
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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)
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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
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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.
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Example
P
Q
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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.
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The gradient of a line joining ( x1 , x2 ) to ( y1 , y2 )
y2 – y1
x2 – x1
2 1
2 1
y y
The gradientx x
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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
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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
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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
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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).
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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
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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
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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
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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
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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)
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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.