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Higher Outcome 1
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Higher Unit 1Higher Unit 1
Distance FormulaThe Midpoint FormulaGradients CollinearityGradients of Perpendicular Lines The Equation of a Straight LineMedian, Altitude & Perpendicular BisectorExam Type Questions
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Higher Outcome 1
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Starter QuestionsStarter Questions
1. Calculate the length of the length AC.
(a)( 1, 2) and ( 5, 10) (b) ( -4, -10) and ( -2,-6)
2. Calculate the coordinates that are halfway between.
6 m
8 m
A
CB
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Higher Outcome 1
Distance FormulaDistance FormulaLength of a straight lineLength of a straight line
A(x1,y1)
B(x2,y2)
x2 – x1
y2 – y1
C
x
y
O
This is just
Pythagoras’ Theorem
2 2 2AB =AC +BC
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Higher Outcome 1
Distance FormulaDistance Formula
The length (distance ) of ANY line can be given by the formula :
2 22 1 2 1tan ( ) ( )dis ceAB y y x x
Just Pythagoras Theorem in
disguise
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Higher Outcome 1
Finding Mid-Point of a lineFinding Mid-Point of a line
A(x1,y1)
B(x2,y2)
x
y
O
1 21 2 , ,2 2
y yx xM
x1 x2
M
y1
y2
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.
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Higher Outcome 1
Straight line FactsStraight line Facts
Y – axis Intercept
2 1
2 1
y - yGradient =
x - x
y = mx + c
Another version of the straight line general formula is:
ax + by + c = 0
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Higher Outcome 1
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m < 0
m > 0
m = 0
x = a
y = c
Sloping left to right up has +ve gradient
Sloping left to right down has -ve gradient
Horizontal line has zero gradient.
Vertical line has undefined gradient.
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Higher Outcome 1
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m = tan θ
m > 0
Lines with the same gradient
means lines are Parallel
The gradient of a line is ALWAYS
equal to the tangent of the angle
made with the line and the positive x-axisθ
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Higher Outcome 1
CollinearityCollinearity
A
C
x
y
O x1 x2
B
Points are said to be collinear
if they lie on the same straight.
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
EF
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Higher Outcome 1
Gradient of perpendicular linesGradient of perpendicular lines
If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2
= -1 When rotated through 90º about the origin A (a, b) → B (-b, a)
-aB(-b,a)
-b
A(a,b)
aO
y
x
- 0
- 0 OA
b bm
a a
- 0-
- - 0 OB
a am
b b
- -1-
OA OB
b a abm m
a b ab
Conversely:
If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.
-b
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Higher Outcome 1
=
The Equation of the Straight LineThe Equation of the Straight Liney – b = m (x - a)
The equation of any line can be found if we know
the gradient and one point on the line.
O
y
x
x - a
P (x, y)
m
A (a, b)y - by - b
x – a
m =y - b
(x – a)m
Gradient,
mPoint (a,
b)
y – b = m ( x – a ) Point on the line ( a, b )
a x
y
b
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Higher Outcome 1
A
B C
D
Apr 18, 2023www.mathsrevision.com 12
AMedian means a line from vertex
to midpoint of the base.
Altitude means a perpendicular line
from a vertex to the base.
B D C
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Higher Outcome 1
Apr 18, 2023www.mathsrevision.com 13
A
B D C
Perpendicular bisector - a line that cuts another line
into two equal parts at an angle of 90o
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Higher Outcome 1
Find the equation of the line which passes through the point
(-1, 3) and is perpendicular to the line with equation 4 1 0x y
Find gradient of given line: 4 1 0 4 1 4x y y x m
Find gradient of perpendicular:1
(using formula 1)1 24 m mm
Find equation:1 3
1 4( 3) 1 4 124 ( 1)
yx y x y
x
4 13 0y x
Typical Exam Typical Exam QuestionsQuestions
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Higher Outcome 1
Find the equation of the straight line which is parallel to the
line with equation
and which passes through the point (2, –1).
Find gradient of given line:
Knowledge: Gradient of parallel lines are the same.
Therefore for line we want to find has gradient
2
3m
Find equation: Using y – b =m(x - a)
3 2 1 0 y x
2 3 5x y
2 2
3 33 2 5 5y x y x m
Typical Exam Typical Exam QuestionsQuestions
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Higher Outcome 1
Find gradient of the line:
1tan
3
Use table of exact values1 1
tan 303
2 ( 1) 3 1
3 3 0 3 3 3m
tanm
Find the size of the angle a° that the line
joining the points A(0, -1) and B(33, 2)
makes with the positive direction of the x-
axis.
Exam Type QuestionsExam Type Questions
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Higher Outcome 1
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: 4 3 5y x
Find mid-point of AB 1, 2
3
4m Find equation of AB
Gradient of AB (perp):4
3m
Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get
3 4 10 0 y x
Exam Type QuestionsExam Type Questions
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Higher Outcome 1
The line AB makes an angle of radians
with
the y-axis, as shown in the diagram.
Find the exact value of the gradient of AB.
Find angle between AB and x-axis:2 3 6
Use table of exact values
3
tanm tan6
m
1
3m
(x and y axes are perpendicular.)
Typical Exam Typical Exam QuestionsQuestions
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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
Find mid-point of BC: 2 1 2 1-(2, 1) Using M ,
2 2
x x y y
Find equation of median AM
Find gradient of median AM
2 1
2 1
-2 Using
-
y ym m
x x
2 5 Using - ( - ) y x y b m x a
Typical Exam Typical Exam QuestionsQuestions
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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:2 1
2 1
-1 Using
2 -
y ym m
x x
Find equation of altitude PS
Find gradient of PS (perpendicular to QR)
1 22 ( 1) m m m
2 3 0 Using ( ) y x y b m x a
Typical Exam Typical Exam QuestionsQuestions
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Higher Outcome 1
The lines and
make angles of a and b with the positive direction of the x-axis, as shown in the diagram.
a) Find the values of a and b
b) Hence find the acute angle between the two given lines.
2m
Find supplement of b 180 135 45
2 4y x 13x y
2 4y x
13x y 1m
Find a° tan 2 63a a
Find b° tan 1 135b b
angle between two lines
Use angle sum triangle = 180°
72°
Typical Exam Typical Exam QuestionsQuestions
45o
72o
63o 135o
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Higher Outcome 1Triangle 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
Find equation of p
2y
Find gradient of p(-2, 2)
Find mid-point of BC
(1, 0) Find gradient of BC
1
2m
0m
Find gradient of q 2m Find equation of q
2 2 y x
Solve p and q simultaneously for intersection
(0, 2)
Exam Type Exam Type QuestionsQuestions p
q
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Higher Outcome 1
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.
7, 2Mid-point AB
Find mid-point AC
(5, 4) Find gradient of AC2
3m
Equation of perp. bisector AC
Gradient AC perp.3
2m 2 3 23y x
Point of intersection (7, 1)
7x Perpendicular bisector AB
Exam Type Exam Type QuestionsQuestions l1
l2
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Higher 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
4, 2Mid-point AB
Equation of median CM using y – b = m(x – a)
Gradient of perpendicular AD
Gradient BC 2m 1
2m
Equation of AD using y – b = m(x – a)
3m Gradient CM (median)
3 14 0 y x
Solve simultaneously for point of intersection(6, -4)
2 2 0y x
Exam TypeExam TypeQuestionsQuestions
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Higher Outcome 1A 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.
2m Gradient AB
Product of gradients
Gradient of median AD
Mid-point BC 3, 1 1
3m Equation AD
1
2m Gradient
BC1
2 12
Solve simultaneously for M, point of intersection
3 6 0y x
Hence AB is perpendicular to BC, so B = 90°
Gradient of median BE
Mid-point AC 2, 3 4
3m Equation
AD3 4 1 0y x
51,
3
Exam Type Exam Type QuestionsQuestions
M