1. Higher Maths 1 1 Straight Line
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Higher Maths Straight Line 2
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Gradient of a Straight Line
The symbol is called ‘delta’ and is used to describe change.
Example
(-2,7) (3,9) x
= 3 – (-2) = 5
B ( x2 , y2 )
Gradient y
y2 – y1
y2 – y1
m = = x
x2 – x1
x2 – x1 A
( x1 , y1 )

3. UNIT OUTCOME SLIDE NOTE y x y x y = x = x y
Higher Maths Straight Line 3
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Horizontal and Vertical Lines
Horizontal
• zero gradient y
• parallel to -axis x
y =
• equations written in the form x
Vertical
(‘undefined’)
• infinite gradient y
• parallel to -axis y
x =
• equations written in the form x

4. UNIT OUTCOME SLIDE NOTE q y q q x q q m m = tan x y x
Higher Maths Straight Line 4
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Gradient and Angle
The symbol is called ‘theta’ and is often used for angles. q y
opp y m
tan q = = = x
adj
opp q x
adj
m = tan q
where is the angle between the line and the positive direction of the -axis. q x

5. x Collinearity y C A Points are said to be collinear
if they lie on the same straight. A C x y O x1 x2 B
The coordinates A,B C are collinear since they lie on the same straight line.
D,E,F are not collinear they do not lie on the same straight line. D E F

6. ( ) , UNIT OUTCOME SLIDE NOTE x2 x1 y2 y1 + + ( x1 , y1 ) M
Higher Maths Straight Line 6
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( x1 , y1 )
Finding the Midpoint
The midpoint M of two coordinates is the point halfway in between and can be found as follows: M
• find the distance between the x-coordinates
• add half of this distance to the first x-coordinate
• do the same again for the y-coordinate
( x2 , y2 )
• write down the midpoint
An alternative method is to find the average of both coordinates:
( )
x2 x1
y2 y1 + + ,
M =
The Midpoint Formula 2 2

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Collinearity
If points lie on the same straight line they are said to be collinear.
Example
Prove that the points P(-6,-5), Q(0,-3) and R(12,1) are collinear.
mPQ =
(-3) – (-5) 2 1
We have shown = = 6 3
mPQ = mQR
0 – (-6)
mQR =
1 – (-3) 4 1
The points are collinear. = =
12 – 0 12 3

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Gradients of Perpendicular Lines A
When two lines are at 90° to each other we say they are perpendicular.
If the lines with gradients m1 and m2 are perpendicular, B P
m1 = –1
mAB = -3
mPQ = 5 m2 5 3
To find the gradient of a perpendicular line: • flip the fraction upside down • change from positive to negative (or negative to positive)

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Different Types of Straight Line Equation
Any straight line can be described by an equation.
The equation of a straight line can be written in several different ways .
Basic Equation (useful for sketching)
y = m x + c
(not the
-intercept!) y
General Equation (always equal to zero)
A x + B y + C = 0
Straight lines are normally described using the General Equation, with as a positive number. A

10. UNIT OUTCOME SLIDE NOTE y x y – b x – a y – b = m ( x – a )
Higher Maths Straight Line 10
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The Alternative Straight Line Equation
Example
Find the equation of the line with gradient and passing through (5,-2). y
( x , y ) 1 2
( a , b ) x
Substitute:
y – b
m =
y – (-2) = ( x – 5 ) 1
x – a 2
y + 2 = ( x – 5 ) 1
y – b = m ( x – a ) 2
2y + 4 = x – 5
for the line with gradient m
passing through (a,b)
x – 2y – 9 = 0

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Straight Lines in Triangles
An altitude is a perpendicular line between a vertex and the opposite side. =
A median is a line between a vertex and the midpoint of the opposite side.
A perpendicular bisector is a perpendicular line from the midpoint of a side. = =

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Finding the Equation of an Altitude
To find the equation of an altitude: B
• Find the gradient of the side it is perpendicular to ( ).
mAB C
• To find the gradient of the altitude, flip the gradient
of AB and change from positive to negative:
maltitude = –1
mAB
• Substitute the gradient
and the point C into
Important
Write final equation in the form
y – b = m ( x – a )
A x + B y + C = 0
with A x positive.

13. ( ) UNIT OUTCOME SLIDE NOTE x2 x1 + y2 y1 + , y – b = m ( x – a ) = =
Higher Maths Straight Line 13
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Finding the Equation of a Median M =
To find the equation of a median: = P
• Find the midpoint of the side it bisects, i.e.
x2 x1 +
y2 y1 + O
( ) ,
M = 2 2
• Calculate the gradient of the median OM.
Important
• Substitute the gradient and either
point on the line (O or M) into
Write answer in the form
A x + B y + C = 0
y – b = m ( x – a )
with A x positive.