Higher Unit 1
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Higher

Outcome 1

Distance Formula
The Midpoint Formula
Gradients
Collinearity
Gradi...
Starter Questions
Outcome 1

www.mathsrevision.com

Higher

2.

1.

Calculate the length of the length AC.
A
6m
Calculate ...
Distance Formula
Length of a straight line
Outcome 1

www.mathsrevision.com

Higher

AB =AC +BC
2

2

2

y
B(x2,y2)

This ...
Distance Formula
www.mathsrevision.com

Higher

Outcome 1

The length (distance ) of ANY line
can be given by the formula ...
Finding Mid-Point of a line
Outcome 1

www.mathsrevision.com

Higher

The mid-point between
2 points is given by
Simply ad...
Straight line Facts
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Higher

Outcome 1

y = mx + c
y2 - y1
Gradient =
x2 - x1

Y – axis
Intercept

A...
Outcome 1

Higher

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Sloping left to right up has +ve gradient
m>0

Sloping left to right down has -v...
Outcome 1

www.mathsrevision.com

Higher

Lines with the same gradient
m>0

means lines are Parallel
The gradient of a lin...
Collinearity
Outcome 1

Higher

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Points are said to be collinear
if they lie on the same straight.

...
Gradient of perpendicular lines
Outcome 1

Higher

→ B (-b, a)

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When rotated through 90º about the ...
The Equation of the Straight Line

y – b = m (x - a)
Outcome 1

Higher

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The equation of any line ca...
Typical Exam Questions
Outcome 1

www.mathsrevision.com

Higher

Find the equation of the line which passes through the po...
Typical Exam Questions
Outcome 1

www.mathsrevision.com

Higher

Find the equation of the straight line which is parallel ...
Outcome 1

Higher

www.mathsrevision.com

Median means a line from vertex
to midpoint of the base.

A

A
B

D

C

D
Altitu...
Outcome 1

www.mathsrevision.com

Higher

Perpendicular bisector means a line from the vertex
that cuts the base in half a...
Exam Type Questions
Outcome 1

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Higher

Find the size of the angle a° that the line
joining the poin...
Exam Type Questions
Outcome 1

Higher

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A and B are the points (–3, –1) and (5, 5).
Find the equatio...
Typical Exam Questions
Outcome 1

www.mathsrevision.com

Higher

π
The line AB makes an angle of
radians with
3
the y-axis...
Typical Exam Questions
Higher

Outcome 1

www.mathsrevision.com

A triangle ABC has vertices A(4, 3), B(6, 1)
and C(–2, –3...
Typical Exam Questions
www.mathsrevision.com

Higher

Outcome 1

P(–4, 5), Q(–2, –2) and R(4, 1) are the vertices
of trian...
www.mathsrevision.com

Higher

Typical Exam
Questions

72o

Outcome 1

The lines y = 2 x + 4

63

and x + y = 13

o

45

o...
Higher

Exam Type
Questions Outcome 1

p
q

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Triangle ABC has vertices
A(–1, 6), B(–3, –2) and C(5, ...
www.mathsrevision.com

Higher

Exam Type
Questions

l2
Outcome 1

l1

Triangle ABC has vertices A(2, 2), B(12, 2) and C(8,...
www.mathsrevision.com

Higher

Exam Type
Questions

Outcome 1

A triangle ABC has vertices A(–4, 1), B(12,3) and C(7, –7)....
Higher

Exam Type
Questions

M

Outcome 1

www.mathsrevision.com

A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,...
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Straight line

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Straight line

  1. 1. Higher Unit 1 www.mathsrevision.com 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 www.mathsrevision.com
  2. 2. Starter Questions Outcome 1 www.mathsrevision.com Higher 2. 1. Calculate the length of the length AC. A 6m Calculate the coordinates that are halfway between. (a) ( 1, 2) and ( 5, 10) (b) www.mathsrevision.com B 8m C ( -4, -10) and ( -2,-6)
  3. 3. Distance Formula Length of a straight line Outcome 1 www.mathsrevision.com 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
  4. 4. Distance Formula www.mathsrevision.com Higher Outcome 1 The length (distance ) of ANY line can be given by the formula : ABdis tan ce = (y2 − y1 ) + (x2 − x1 ) Just Pythagoras Theorem in disguise
  5. 5. Finding Mid-Point of a line Outcome 1 www.mathsrevision.com Higher 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 y2 A(x1,y1) M y1 O  x1 + x 2 y1 + y 2  M = , ,÷ 2  2  B(x2,y2) x1 x2 x
  6. 6. Straight line Facts www.mathsrevision.com 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
  7. 7. Outcome 1 Higher www.mathsrevision.com 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 www.mathsrevision.com Feb 2, 2014 7
  8. 8. Outcome 1 www.mathsrevision.com 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 www.mathsrevision.com Feb 2, 2014 8
  9. 9. Collinearity Outcome 1 Higher www.mathsrevision.com 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
  10. 10. Gradient of perpendicular lines Outcome 1 Higher → B (-b, a) www.mathsrevision.com When rotated through 90º about the origin A (a, b) mOB a-0 a = =-b - 0 b B(-b,a) y -a -b -b A(a,b) O mOA ×mOB = a mOA b-0 b = = a -0 a x b a ab × == -1 a -b ab If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1 Conversely: If m1 × m2 = -1 then the two lines with gradients m 1 and m2 are perpendicular.
  11. 11. The Equation of the Straight Line y – b = m (x - a) Outcome 1 Higher www.mathsrevision.com The equation of any line can be found if we know the gradient and one point on the line. y P (x, y) y m b O A (a, b) a x–a y-b m= (x – a) y -- b b x x Gradient, y–b=m(x–a) m Point (a, b) Point on the line ( a, b )
  12. 12. Typical Exam Questions Outcome 1 www.mathsrevision.com 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 y – b = m(x – a) y – 3 = ¼ (x –(-1)) 4y – 12 = x + 1 4 y − x − 13 = 0
  13. 13. Typical Exam Questions Outcome 1 www.mathsrevision.com Higher Find the equation of the straight line which is parallel to the line 2 x + 3with 5 y = 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 = − Gradient of parallel line is same: Find equation: M = -2/3 y – b = m(x – a) y – (-1) = -2/3 (x – 2) 3y + 3 = -2x + 4 3 y + 2x =1 2 3
  14. 14. Outcome 1 Higher www.mathsrevision.com Median means a line from vertex to midpoint of the base. A A B D C D Altitude means a perpendicular line B C from a vertex to the base. www.mathsrevision.com Feb 2, 2014 14
  15. 15. Outcome 1 www.mathsrevision.com Higher Perpendicular bisector means a line from the vertex that cuts the base in half and at right angles. A B D www.mathsrevision.com C Feb 2, 2014 15
  16. 16. Exam Type Questions Outcome 1 www.mathsrevision.com 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 m = tan θ tan θ = Use table of exact values 1 3 θ = tan −1 1 θ = 30° 3 1 3
  17. 17. Exam Type Questions Outcome 1 Higher www.mathsrevision.com 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 3 Find equation of AB 4 y = 3 x + 5 Find gradient of the AB: m = 4 Find mid-point of AB (1,2) 4 Gradient of AB (perp): m = − 3 Use gradient and mid-point to obtain perpendicular bisector AB 4x + 3y = 10
  18. 18. Typical Exam Questions Outcome 1 www.mathsrevision.com Higher π 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. π π π − = Find angle between AB and x-axis: 2 3 6 π m = tan θ m = tan 6 Use table of exact values 1 m= 3 (x and y axes are perpendicular.)
  19. 19. Typical Exam Questions Higher Outcome 1 www.mathsrevision.com 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 A.  x2 - x1 y2 − y1  , 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 )
  20. 20. Typical Exam Questions www.mathsrevision.com 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 = 1 y -y 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 )
  21. 21. www.mathsrevision.com Higher Typical Exam Questions 72o Outcome 1 The lines y = 2 x + 4 63 and x + y = 13 o 45 o 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. y = 2x + 4 m=2 Find a° tan a° = 2 → a = 63° 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°
  22. 22. Higher Exam Type Questions Outcome 1 p q www.mathsrevision.com 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 m=0 1 Find gradient of BC 2 y = −2 x + 2 Find equation of q Solve p and q simultaneously for intersection (0, 2) m=
  23. 23. www.mathsrevision.com 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)
  24. 24. www.mathsrevision.com 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 ) Equation of median CM Gradient BC Equation of AD m=2 Gradient CM (median) m = −3 y + 3x = 14 Gradient of perpendicular AD 2y + x + 2 = 0 Solve simultaneously for point of intersection (6, -4) 1 m=− 2
  25. 25. Higher Exam Type Questions M Outcome 1 www.mathsrevision.com 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 ) Gradient of median BE m = − Equation AD 3 y + 4 x + 1 = 0 Mid-point AC ( 3 5  Solve simultaneously for M, point of intersection  1, − ÷ 3  Mid-point BC ( 3, − 1) Gradient of median AD m =

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