1. Straight Line Applications 1.1
Prior Knowledge
Distance Formula
The Midpoint Formula
m = tan θ
Gradients of Perpendicular Lines
Median, Altitude & Perpendicular Bisector
Collinearity
Mindmap
Exam Type Questions

2. Circle Vector Logs & Exponentials Differentiation
Recurrence Relations
Vector
Where is
Straight Line Theory
Used in Higher
Logs & Exponentials
Differentiation

3. Parallel lines have same gradient
Distance between
2 points
m < 0
m = undefined
Terminology
Median – midpoint
Bisector – midpoint
Perpendicular – Right Angled
Altitude – right angled
m1.m2 = -1
m > 0
m = 0
Possible values for gradient
Form for finding
line equation
y – b = m(x - a)
(a,b) = point on line
Straight Line
y = mx + c
Parallel lines have same gradient
m = gradient
c = y intercept (0,c)
For Perpendicular lines the following is true.
m1.m2 = -1 θ
m = tan θ

4. Another version of the straight line formula is
Straight Line Facts
y = mx + c
Y – axis Intercept
Another version of the straight line formula is
ax + by + c = 0

5. Find the gradient and y – intercept for equations.
Questions
Find the gradient and y – intercept for equations.
(a) 4x + 2y + 10 = 0
m = c = -5
(b) 3x - 5y + 1 = 0
m = 3/5 c = 1/5
(c) 4 - x – 3y = 0
m = - 1/3 c = 4/3
Demo

6. The Equation of the Straight Line y – b = m (x - a)
Demo
The equation of any line can be found if we know
the gradient and one point on the line. O y x
P (x, y) m
A (a, b)
y - b y m
m = =
y - b
(x – a)
y - b b
x - a
x – a
Gradient, m a x
Point (a, b)
y – b = m ( x – a )
Point on the line ( a, b )

7. Gradient Facts 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
24-Apr-17

8. Gradient Facts Lines with the same gradient means lines are Parallel
24-Apr-17

9. Straight Line Theory

10. Straight Line Theory

11. Straight Line Theory

12. Straight Line Theory

13. Typical Exam Questions
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
Find equation: Using y – b =m(x - a)

14. HHM Book Practice
HHM Ex1A
HHM Ex1B Q4 , Q5
HHM Ex1E
HHM Ex1G Q1 , Q2

15. Distance Formula Length of a straight linex y O
B(x2,y2)
This is just
Pythagoras’ Theorem
y2 – y1
A(x1,y1) C
x2 – x1

16. Distance Formula
The length (distance ) of ANY line can be given by the formula :
Just Pythagoras Theorem in disguise
Demo

17. 2√26
2√26
Isosceles Triangle !
8√2
Demo

18. HHM Book Practice
HHM Ex12B

19. x Mid-Point of a line y B(x2,y2) A(x1,y1) The mid-point (Median)
between 2 points is given by x y O
B(x2,y2) y2
Simply add both x coordinates together and divide by 2.
Then do the same with the y coordinates. M
A(x1,y1) y1 x1 x2
Online Demo

21. HHM Book Practice
Use Show me boards

22. Straight Line Facts m = tan θ m = Demo tan θ =
The gradient of a line is ALWAYS
equal to the tangent of the angle
made with the line and the positive x-axis θ
m = tan θ
0o ≤ θ < 180o θ O A H
m = V H O A = O A
tan θ =
Demo
24-Apr-17

23. m = tan θ
60o
m = tan 60o = √3

24. m = tan θ
y = -2x
m = tan θ
θ = tan-1 (-2)
θ = 180 – 63.4
θ = 116.6o

25. Exam Type Questions 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.
Find gradient of the line:
Use
Use table of exact values

26. Typical Exam Questions
The line AB makes an angle of 60o with
the y-axis, as shown in the diagram.
Find the exact value of the gradient of AB.
60o
Find angle between AB and x-axis:
(x and y axes are perpendicular.)
Use
Use table of exact values

27. Typical Exam Questions
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.
Find gradient of
Find a°
Find gradient of
Find b°
Find supplement of b
72°
Use angle sum triangle = 180°
angle between two lines

28. HHM Book Practice
HHM Ex1A Q9 , Q10
HHM Ex1B Q6 , Q7 , Q10
HHM Ex1D Q7

29. Gradient of perpendicular lines
Investigation
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) y
B(-b,a) -a
A(a,b) -b O a x -b
Demo
Conversely:
If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.

30. HHM Book Practice
HHM Ex1D
HHM Ex1F
HHM Ex1G Q3 – Q5

31. Terminology A B C D A B C D Demo Demo Median means a line
from a vertex to
the midpoint of the base. B C D A B C D
Demo
Altitude means a perpendicular line
from a vertex to the base.
Demo
24-Apr-17

32. Terminology
Perpendicular bisector - a line that cuts another line
into two equal parts at an angle of 90o A B C D
Demo
24-Apr-17

33. Common Straight Strategies for Exam Questions
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.

34. Common Straight Strategies for Exam Questions
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.

35. Demo Any number of lines are said to be concurrent
if there is a point through which they all pass.
For three lines to be concurrent,
they must all pass through a single point.
Demo

36. HHM Book Practice
HHM Ex1I
HHM Ex1K
HHM Ex1M
HHM Ex1N

37. 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

38. they have a point in common Q
Ratio
DPQ:DPQ
1:2
Straight Line Theory
DPQ=2√2
DPQ=√2
Since mPQ = mQR and
they have a point in common Q
PQR are collinear.

39. HHM Book Practice
HHM Ex1B Q1 – Q3
HHM Ex1O

40. Typical Exam Questions
Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation
Find gradient of given line:
Find gradient of perpendicular:
Find equation:

41. Find gradient of the AB:
Exam Type Questions
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:
Find equation of AB
Find mid-point of AB
Gradient of AB (perp):
Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get

42. Typical Exam Questions
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:
Find gradient of median AM
Find equation of median AM

43. Typical Exam Questions
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:
Find gradient of PS (perpendicular to QR)
Find equation of altitude PS

44. Solve p and q simultaneously for intersection
Exam Type Questions p
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. q
Find mid-point of AB
(-2, 2)
Find gradient of p
Find equation of p
Find mid-point of BC
(1, 0)
Find gradient of BC
Find gradient of q
Find equation of q
Solve p and q simultaneously for intersection
(0, 2)

45. Exam Type Questions l2 l1 (5, 4) (7, 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.
Mid-point AB
Perpendicular bisector AB
Find mid-point AC
(5, 4)
Find gradient of AC
Gradient AC perp.
Equation of perp. bisector AC
Point of intersection
(7, 1)

46. Gradient of perpendicular AD
Exam Type Questions
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
Gradient CM (median)
Equation of median CM using y – b = m(x – a)
Gradient BC
Gradient of perpendicular AD
Equation of AD using y – b = m(x – a)
Solve simultaneously for point of intersection
(6, -4)

47. Hence AB is perpendicular to BC, so B = 90°
Exam Type Questions M
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
Gradient BC
Product of gradients
Hence AB is perpendicular to BC, so B = 90°
Mid-point BC
Gradient of median AD
Equation AD
Mid-point AC
Gradient of median BE
Equation AD
Solve simultaneously for M, point of intersection

48. Parallel lines have same gradient
Distance between
2 points
m < 0
m = undefined
Terminology
Median – midpoint
Bisector – midpoint
Perpendicular – Right Angled
Altitude – right angled
m1.m2 = -1
m > 0
m = 0
Possible values for gradient
Form for finding
line equation
y – b = m(x - a)
(a,b) = point on line
Straight Line
y = mx + c
Parallel lines have same gradient
m = gradient
c = y intercept (0,c)
For Perpendicular lines the following is true.
m1.m2 = -1 θ
m = tan θ

49. Are you on Target ! Update you log book
Make sure you complete and correct
ALL of the Straight Line questions in
the past paper booklet.