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Syllabus Topic: Coordinate Geometry
Title: Plotting Points, Gradient, Parallel and Perpendicular Lines, Distance, Midpoint, Equation of a Line
Plotting Points
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A point is defined as an ordered pair or coordinate, (��, ��), such that �� is determined by its horizontal distance from the origin and �� is determined by its vertical distance from the origin.
Consider the arrow diagram below:
�� ��
Rule: �� = 5��
1 2 3
V V V
5
10 15
input output
To show this information on a graph, we will plot the following points: • (1,5)
• (2,10)
• (3,15)
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Title: Graph showing the relation �� = 5��.
output
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Worked Example
Plot the following points on the same grid:
(a) ��(2,3) (e) ��(−3, −4) (b)��(0,5) (f) ��(4, −3) (c) ��(������������) (g) ��(1, −2) (d)��(−5,0) (h) ��(0, −1)
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Consider the point ��(2,3)
��-coordinate ��-coordinate
Note: Points are always written of the form (��, ��).
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The number 2 is called the ��-coordinate.
The number 3 is called the ��-coordinate.
��
��
��
��
��
��
��
��
��
��
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Points to Note:
• Plan Ahead: Notice that on the ��-axis, you need to include values from −5 to 4 and on the ��-axis, you need to include values from −4 to 5. Essentially, you should
have 5 units on each side of the origin for this graph.
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• Ensure to occupy 80% of the graph page when drawing graphs.
• Use a small “×” to plot the points.
• Points are normally represented by capital letters and lines are represented by common letters.
• The origin has coordinates (0,0).
Gradient
➢ Another word for gradient is “slope” or “steepness”.
➢ The gradient of a straight line tells us how steep a line is, therefore, the bigger the gradient, the steeper the line.
Exercise: Which line represents each gradient below?
Gradient
Representation of Line
Positive
R
Negative
Zero
Infinite
(undefined)
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Solution:
Gradient
Representation of Line
Positive
Negative
Zero
R
Infinite
(undefined)
A
Parallel Lines
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Parallel lines are a fixed distance apart and will never meet, no matter how long they are extended.
��1
��2
Lines that are parallel have the same gradient.
Therefore, for parallel lines ���� = ����.
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Perpendicular Lines
Two lines are perpendicular if one is at right angles to each other. In other words, if the two lines cross and the angle between them is 90°.
��2
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��1
If two lines are perpendicular, then their gradients will multiply together to give -1. Therefore, for perpendicular lines, ���� = −������.
Gradient Formula
The gradient can be calculated as the change in �� over the change in ��.
Gradient = ��������
������
Gradient = ����
����
Gradient = ����−����
����−����
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Worked Example
��
��
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��
��
(a) What are the coordinates of �� and ��?
(b)What is the gradient of the line ����?
Solution:
(a) The coordinates of point �� is ��(10,20).
The coordinates of point �� is ��(45,35).
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(b)Title: Graph showing the line ����. ��
Note: Line ���� is the same as line ����.
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��
��
��
Consider ��(10,20) and ��(45,35)
(��1, ��1) (��2, ��2)
Note: It does not matter which point is labelled as (��1, ��1) and (��2, ��2).
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Gradient, m = ��2−��1
��2−��1
= 35−20
45−10
= 15
35
Note: Leave answer as a fraction.
= 37
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Distance/Length of a line
To calculate the distance between two points in a straight line, we use the distance formula below which is an application of Pythagoras’ Theorem.
Distance Formula:
Distance = √(���� − ����)�� + (���� − ����)��
Relation of the distance formula to Pythagoras’ Theorem
Recall:
��
��
��
Pythagoras’ Theorem states that: ���� = ���� + ����
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Consider the gradient of the line ����.
��
��
��2
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��
��1
��
��
��1 ��2
Using Pythagoras’ Theorem to find the distance of the line ����, we get: ��2 = ��2 + ��2
��2 = (��2 − ��1)2 + (��2 − ��1)2
�� = √(��2 − ��1)2 + (��2 − ��1)2
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Worked Example
��
��(3,8)
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��(12,3)
��(0,0)
��
(a) Find the gradient of the line ����.
(b)State which line has a larger gradient between ���� and ����? (c) Find the length of the line ����.
(d)Find the midpoint of ����.
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Solution:
(a) The gradient of the line ���� is calculated as follows:
Consider ��(3 , 8) and ��(12 , 3)
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(��1, ��1) (��2, ��2)
Gradient of ����, �� = ��2−��1
��2−��1
= 3−8
12−3
= −59
OR
Consider ��(12, 3) and ��(3 , 8) (��1, ��1) (��2, ��2)
Gradient of ����, m = ��2−��1
��2−��1
= 8−3
3−12
= 5−9
Note:
−��
��=��−��= −����
Note: It does not matter which point is labelled as (��1, ��1) and (��2, ��2).
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(b)Title: Graph showing line ���� and ����
��
��(3,8)
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��(12,3)
��(0,0)
��
Looking at the graph, it can be deduced that the line ���� is steeper than ����. Hence, line ���� has a larger gradient.
(c) Consider ��(3 , 8) and ��(12 , 3)
(��1, ��1) (��2, ��2)
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The distance of line ���� is calculated as follows:
Distance of line ���� = √(��2 − ��1)2 + (��2 − ��1)2
= √(12 − 3)2 + (3 − 8)2
Note: Ensure to put your
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= √92 + (−5)2
= √81 + 25
= √106
= 10.3 units (to 3 significant figures) (d)To calculate the midpoint, the following formula is used:
Midpoint Formula:
Midpoint = (����+����
��,����+����
��)
Consider ��(0 , 0) and ��(12 , 3)
(��1, ��1) (��2, ��2)
Midpoint = (��1+��2
2,��1+��2
2)
= (0+12
2,0+3
2)
= (122,32)
= (6,32)
units when calculating distance. If units are not given, just write “units” after your answer.
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Worked Example 2
��
��
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��
��
��
(a) Find the gradient of the line ����.
(b)Find the distance ����.
(c) Find the midpoint, ��, of ����.
(d)Find the gradient of ����.
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Solution:
��
��(6, 18)
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×
��(2, 4)
��
��(4, 11)
��(14, 10)
(a) Points are ��(2, 4) and ��(6, 18). The gradient of ���� is as follows:
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Gradient = ��2−��1
��2−��1
Gradient = 18−4
6−2
Gradient = 144
Gradient = 3.5
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(b)Points are ��(2, 4) and ��(6, 18).
Distance ���� = √(��2 − ��1)2 + (��2 − ��1)2
Distance ���� = √(6 − 2)2 + (18 − 4)2
Distance ���� = √(4)2 + (14)2
Distance ���� = √16 + 196
Distance ���� = √212
Distance ���� = 14.6 units
(c) Midpoint, �� = (��1+��2
2,��1+��2
2)
Midpoint, �� = (2+6
2,4+18
2)
Midpoint, �� = (82,222)
Midpoint, �� = (4, 11)
(d)Points are ��(4, 11) and ��(14, 10).
Gradient of ���� =��2−��1
��2−��1
Gradient of ���� =10−11
14−4
Gradient of ���� =−1
10
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Equation of a line
The general form for the equation of a line is
�� = ���� + ��
where �� = gradient of the line
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�� = ��-intercept
Worked Example
��
��(4,11)
��(0,3)
��
Find the equation of the line ����.
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Solution:
Consider ��(0 , 3) and ��(4, 11)
(��1, ��1) (��2, ��2)
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Gradient of ����, �� = ��2−��1
��2−��1
= 11−3
4−0
= 84
= 2
From the graph, it can be seen that the line cuts the ��-axis at 2. Hence, �� = 3. ∴ The equation of the line ���� is �� = 2�� + 3.
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Exercise
��
��
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��
��
��
Which of the following equations corresponds to the lines ��, �� and ��? �� =12�� + 4
�� = −1�� + 12
�� =12�� + 8
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Solution:
Note
��
��
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��
��
��
Equation of line: �� = ���� + �� If the constants �� and �� change, the line replicates this change. This can be seen in the following table:
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Equation of line
Representation of line
�� = 2�� + 4
UNAR
R�� = −2�� + 4
JAM
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�� = 3�� + 10
UNAR
R�� = 3�� + 20
JAM
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�� = 3�� − 1
UNAR
R�� = 3�� − 15
JAM
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Worked Example 1
A line has a gradient of 12and passes through the point (2,3). Find the equation of the line. Solution:
METHOD 1:
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• Substitute into �� = ���� + ��
• Find ��.
• Write equation.
Substitute point (2,3) and gradient = 12into �� = ���� + ��, 3 =12(2) + ��
3 = 1 + ��
�� = 3 − 1
�� = 2
So, �� = 2.
∴ The equation of the line is �� =12�� + 2,
in the form �� = ���� + ��, where �� =12and �� = 2.
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METHOD 2:
• Use �� − ���� = ��(�� − ����)
�� =��2−��1 ��2−��1
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Rearranging this equation and eliminating one of the points gives, �� − ��1 = ��(�� − ��1)
Substitute point (2,3) and gradient = 12into �� − ��1 = ��(�� − ��1), �� − 3 =12(�� − 2)
�� − 3 =12�� − 1
�� =12�� − 1 + 3
�� =12�� + 2
∴ The equation of the line is �� =12�� + 2,
in the form �� = ���� + ��, where �� =12and �� = 2.
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Worked Example 2
Find the equation of the line passing through the points ��(1, 10) and ��(5, −4).
Solution:
The points are ��(1, 10) and ��(5, −4).
�� =��2−��1 ��2−��1
�� =−4−10 5−1
�� =−14 4
�� = −72
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Substituting �� = −72and point (1, 10) into �� = ���� + �� gives, 10 = −72(1) + ��
10 = −72+ ��
�� = 10 +72
�� =272
∴ The equation of the line is �� = −72�� +272,
in the form �� = ���� + ��, where �� = −72and �� =272.
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Alternatively,
Substituting �� = −72and point (1, 10) into �� − ��1 = ��(�� − ��1) gives, �� − 10 = −72(�� − 1)
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�� − 10 = −72�� +72
�� = −72�� +72+ 10
�� = −72�� +272
∴ The equation of the line is �� = −72�� +272,
in the form �� = ���� + ��, where �� = −72and �� =272.
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Finding the equation of a perpendicular line
Question:
��
��(3, 12)
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��(1, 2)
��(4, 0)
��
Find:
(a) Find the gradient of the line ����.
(b)Find the equation of the line perpendicular to ���� but passing through point ��.
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Solution:
(a) Points are ��(1, 2) and ��(3, 12).
Gradient of ���� =��2−��1
��2−��1
Gradient of ���� =12−2
3−1
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Gradient of ���� =102
Gradient of ���� = 5
(b)��1 = 5
So, the gradient of the perpendicular line, ��2 = −1��1
So, the gradient of the perpendicular line, ��2 = −15
Substituting point (4, 0) and gradient = −15into �� − ��1 = ��(�� − ��1)
�� − 0 = −15(�� − 4)
�� = −15�� +45
∴ The equation of the line is �� = −15�� +45,
in the form �� = ���� + ��, where �� = −15and �� =45.