Higher Unit 1
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Higher
Outcome 1
Distance Formula
The Midpoint Formula
Gradients
Collinearity
Gradients of Perpendicular Lines
The Equation of a Straight Line
Median, Altitude & Perpendicular Bisector
Exam Type Questions
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Starter Questions
Outcome 1
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Higher
1.
Calculate the length of the length AC.
A
6m
2.
Calculate the coordinates
that are halfway between.
(a) ( 1, 2) and ( 5, 10)
(b)
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B
8m
C
( -4, -10) and ( -2,-6)
Distance Formula
Length of a straight line
Outcome 1
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Higher
AB =AC +BC
2
2
2
y
B(x2,y2)
This is just
y2 – y1
A(x1,y1)
Pythagoras’ Theorem
x2 – x1
O
C
x
Distance Formula
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Higher
Outcome 1
The length (distance ) of ANY line
can be given by the formula :
ABdis tan ce  (y2  y1 )2  (x2  x1 )2
Just
Pythagoras
Theorem in
disguise
Finding Mid-Point of a line
Outcome 1
Higher
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The mid-point between
2 points is given by
Simply add both x
coordinates together
and divide by 2.
Then do the same with
the y coordinates.
y
M
A(x1,y1)
y1
O
 x1  x 2 y 1  y 2 
M 
,
,
2
 2

B(x2,y2)
y2
x1
x2
x
Straight line Facts
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Higher
Outcome 1
y = mx + c
y2 - y1
Gradient =
x2 - x1
Y – axis
Intercept
Another version of the straight line general formula is:
ax + by + c = 0
Outcome 1
Higher
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Sloping left to right up has +ve gradient
m>0
Sloping left to right down has -ve gradient
m<0
Horizontal line has zero gradient.
m=0
y=c
Vertical line has undefined gradient.
x=a
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7
Outcome 1
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Higher
Lines with the same gradient
m>0
means lines are Parallel
The gradient of a line is ALWAYS
equal to the tangent of the angle
θ
made with the line and the positive x-axis
m = tan θ
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8
Collinearity
Outcome 1
Higher
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Points are said to be collinear
if they lie on the same straight.
y
C
The coordinates A,B C are
collinear since they lie on
the same straight line.
B
D,E,F are not collinear they
do not lie on the same
straight line.
A
E
F
D
O
x1
x2
x
Gradient of perpendicular lines
Outcome 1
Higher
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If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1
When rotated through 90º about the origin A (a, b)
y
B(-b,a)
mOB
a-0
a

-b - 0
b
-a
-b
-b
mOA  mOB
→ B (-b, a)
A(a,b)
O
mOA
a
b-0 b


a-0 a
x
b a
ab
 
 -1
a -b
ab
Conversely:
If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.
The Equation of the Straight Line
y – b = m (x - a)
Outcome 1
Higher
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The equation of any line can be found if we know
the gradient and one point on the line.
y
y
P (x, y)
m
b
O
m=
y -- bb
y-b
(x – a)
A (a, b)
xx –- aa
a
x
x
Gradient, m Point (a,
y–b=m(x–a)
b)
Point on the line ( a, b )
Outcome 1
Higher
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Median means a line from vertex
to midpoint of the base.
A
A
B
D
D
C
Altitude means a perpendicular line
B
C
from a vertex to the base.
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12
Outcome 1
Higher
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Perpendicular bisector - a line that cuts another line
into two equal parts at an angle of 90o
A
B
D
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C
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13
Typical Exam Questions
Outcome 1
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Higher
Find the equation of the line which passes through the point (-1, 3)
and is perpendicular to the line with equation 4 x  y  1  0
Find gradient of given line:
4 x  y  1  0  y  4 x  1  m  4
Find gradient of perpendicular:
Find equation:
m
1
(using formula m  m  1)
1
2
4
1
y 3

 x  1  4( y  3)  x  1  4 y  12
4 x ( 1)
4 y  x  13  0
Typical Exam Questions
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Higher
Outcome 1
Find the equation of the straight line which is parallel to the line
2 x  3 y  5 with equation
and which passes through the point (2, –1).
Find gradient of given line:
2
3
3 y  2 x  5  y   x  5  m  
2
3
Knowledge: Gradient of parallel lines are the same.
Therefore for line we want to find has gradient
Find equation: Using y – b =m(x - a)
3y  2x 1  0
m
2
3
Exam Type Questions
Outcome 1
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Higher
Find the size of the angle a° that the line
joining the points A(0, -1) and B(33, 2)
makes with the positive direction of the x-axis.
2  (1)
3


Find gradient of the line: m 
3 3 0
3 3
Use m  tan 
tan  
Use table of exact values
1
3
  tan
1
1
  30
3
1
3
Exam Type Questions
Outcome 1
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Higher
A and B are the points (–3, –1) and (5, 5).
Find the equation of
a) the line AB.
b) the perpendicular bisector of AB
Find gradient of the AB: m 
Find mid-point of AB
1, 2 
3
4
Find equation of AB 4 y  3 x  5
4
Gradient of AB (perp): m  
3
Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular
bisector of AB we get
3 y  4 x  10  0
Typical Exam Questions
Outcome 1
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Higher

The line AB makes an angle of
radians with
3
the y-axis, as shown in the diagram.
Find the exact value of the gradient of AB.
Find angle between AB and x-axis:
Use m  tan 
m  tan
Use table of exact values
  
 
2 3 6

6
1
m
3
(x and y axes are perpendicular.)
Typical Exam Questions
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Higher
Outcome 1
A triangle ABC has vertices A(4, 3), B(6, 1)
and C(–2, –3) as shown in the diagram.
Find the equation of AM, the median from B to C
 x1  x2 y1  y2 
,
Find mid-point of BC: (2, 1) Using M 

2
2


Find gradient of median AM m  2 Using m 
y2 - y1
x2 - x1
Find equation of median AM y  2 x  5 Using y - b  m( x - a)
Typical Exam Questions
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Higher
Outcome 1
P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices
of triangle PQR as shown in the diagram.
Find the equation of PS, the altitude from P.
Find gradient of QR: m 
y -y
1
Using m  2 1
2
x2 - x1
Find gradient of PS (perpendicular to QR)
m   2 (m1  m2  1)
Find equation of altitude PS
y  2x  3  0
Using y  b  m( x  a )
Higher
Typical Exam
Questions
Outcome 1
The lines y  2 x  4
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72o
63o
and x  y  13
45o
135o
make angles of a and b with the positive direction of the xaxis, as shown in the diagram.
a) Find the values of a and b
b) Hence find the acute angle between the two given lines.
Find gradient of
y  2x  4
m2
Find a° tan a  2  a  63
Find gradient of
x  y  13
m  1
Find b° tan b  1  b  135
Find supplement of b
 180  135  45
Use angle sum triangle = 180° angle between two lines
72°
Exam Type
Questions Outcome 1
Higher
p
q
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Triangle ABC has vertices
A(–1, 6), B(–3, –2) and C(5, 2)
Find:
a) the equation of the line p, the median from C of triangle ABC.
b) the equation of the line q, the perpendicular bisector of BC.
c) the co-ordinates of the point of intersection of the lines p and q.
Find mid-point of AB (-2, 2)
Find equation of p
y2
Find mid-point of BC
(1, 0)
Find gradient of q
m  2
Find gradient of p
Find gradient of BC
Find equation of q
Solve p and q simultaneously for intersection (0, 2)
m0
m
1
2
y  2 x  2
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Higher
Exam Type
Questions
l2
Outcome 1
l1
Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6).
a) Write down the equation of l1, the perpendicular bisector of AB
b) Find the equation of l2, the perpendicular bisector of AC.
c) Find the point of intersection of lines l1 and l2.
Mid-point AB
 7, 2 
Find mid-point AC
Gradient AC perp.
Point of intersection
Perpendicular bisector AB
x7
2
(5, 4) Find gradient of AC m 
3
3
m
Equation of perp. bisector AC 2 y  3 x  23
2
(7, 1)
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Higher
Exam Type
Questions
Outcome 1
A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7).
a) Find the equation of the median CM.
b) Find the equation of the altitude AD.
c) Find the co-ordinates of the point of intersection of CM and AD
Mid-point AB
 4, 2 
Gradient CM (median)
Equation of median CM using y – b = m(x – a)
Gradient BC
m2
m  3
y  3x  14  0
Gradient of perpendicular AD
Equation of AD using y – b = m(x – a)
2y  x  2  0
Solve simultaneously for point of intersection (6,
-4)
m
1
2
Exam Type
Questions
Higher
M
Outcome 1
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A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3).
a) Show that the triangle ABC is right angled at B.
b) The medians AD and BE intersect at M.
i) Find the equations of AD and BE. ii) Find find the co-ordinates of M.
Gradient AB
m2
Product of gradients
Gradient BC
2  
1
 1
2
1
m
2
Hence AB is perpendicular to BC, so B = 90°
1
Equation AD 3 y  x  6  0
3
4
2,

3
m


Equation AD 3 y  4 x  1  0
Mid-point AC 
 Gradient of median BE
3
5

1,

Solve simultaneously for M, point of intersection 

3

Mid-point BC  3, 1 Gradient of median AD m 