1. Higher
Linear Relationships
Lesson 7

2. (-0.8, 0) (-1,0.5) (0,2) (2,3) I don’t know
Linear Relationships
Which of the following points does not lie on the line 2y + 5x – 4 = 0? 6
(-0.8, 0)
(-1,0.5)
(0,2)
(2,3)
I don’t know 1 2 3 4 5 6

3. Linear Relationships
1 3y = 4x y = 3x y + 3x = x + 3y = 2 6
Lines 1 and 2 are perpendicular
Lines 1 and 4 are parallel
Lines 2 and 4 are perpendicular
Lines 2 and 3 are parallel
I don’t know 1 2 3 4 5 6

4. The gradient is 0.3 and the y intercept is 1.5
Linear Relationships
A straight line has equation 10y = 3x Which of the following is true? 6
The gradient is 0.3 and the y intercept is 1.5
The gradient is 3 and the y intercept is 15
The gradient is 15 and the y intercept is 3
The gradient is 1.5 and the y intercept is 0.3
I don’t know 1 2 3 4 5 6

5. To find where two lines meet we can use simultaneous equations
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
Intersecting Lines
To find where two lines meet we can use simultaneous equations
For this reason it is useful to write the equations in the form
Ax + By = C
Tuesday, 03 July 2018

6. Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2).
Example 1
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
Intersecting Lines
Example 1
Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2).
The median from Q meets the altitude
from P at point K.
Draw a diagram
P(-3,6)
Q(5,12)
R(5,2) K
Find the equations of the median and altitude.
Hence find the co-ordinates of K.
Tuesday, 03 July 2018

7. Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2).
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
Intersecting Lines
Triangle PQR has vertices P(-3,6) , Q(5,12) & R(5,2).
Find the equations of the median and altitude.
To find the equation of the median QK
y - b = m(x - a)
y - 4 = 2(x - 1)
Find the midpoint of PR
Gradient of median QK = (12 - 4)
(5-1)
mQK = 2
y - 4 = 2x - 2
Using
P = (-3, 6)
R = (5, 2)
(1,4). x
-3 + 5 2 y
6 + 2 2
2x - y = -2
Q(5,12) K
P(-3,6)
R(5,2)
Midpoint of PR , K, is (1,4)
To find the equation of the altitude PK
Find gradient QR
mQR = (12 - 2) / (5 - 5)
= 10 / 0
= undefined
y - b = m( x - a)
y - 6 = 0(x + 3)
mPK
PK is zero
y - 6 = 0
y = 6

8. To find the point of intersection of K
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
Intersecting Lines
To find the point of intersection of K
2x - y = (A)
y = (B)
Substitute y = 6 in the top equation
2x - 6 = -2
P(-3,6)
Q(5,12)
R(5,2) K
2x = 4
x = 2
Hence K is the point (2,6)
Tuesday, 03 July 2018

9. Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
Example 2
The points E(-1,1), F(3,3) and G(6,2) all lie on the circumference of a circle.
Find the equations of the perpendicular bisectors of EF and FG.
E(-1,1)
F(3,3)
G(6,2) C
Hence find the co-ordinates of the centre of the circle, C.
Diagram will be something like

10. A B Higher Mathematics Linear Relationships Unit 1 Outcome 1
Find the equations of the perpendicular bisectors of EF and FG.
Find Midpoint of EF
Find Midpoint of FG
Using
E = (-1, 1)
F = (3, 3)
Using
F = (3, 3)
G = (6, 2) x
3 + 6 2 y
3 + 2 2 x
-1 + 3 2 y
1 + 3 2
E(-1,1)
F(3,3)
G(6,2) C
Midpoint of FG is B (4.5, 2.5)
Midpoint of EF is A (1,2)
mFG = (2-3)/(6-3)
= -1/3 B
mEF = (3-1)/(3+1)
= 1/2 A
Perpendicular m = -2
Perpendicular m = 3
y - b = m(x - a)
y - b = m(x – a)
y - 2 = -2( x - 1)
y = 3( x - 4.5)
y - 2 = -2x + 2
y = 3x
2x + y = 4
3x - y = 11 A B

11. Linear Relationships
Unit 1 Outcome 1
Finding where the bisectors meet gives us the centre of the circle
Solving
2x + y = (A)
3x - y = (B)
5x = 15
E(-1,1)
F(3,3)
G(6,2) C
x = 3
Substituting x = 3 into A
2x + y = 4
6 + y = 4
y = -2
Hence centre of circle at (3,-2)

12. Intersecting Straight Lines using simultaneous equations
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
The Equation of the straight line
Intersecting Straight Lines using simultaneous equations
Homework
Page xx Exercise A
Question xxxx
Classwork
Page 11 Exercise 6
Complete
Tuesday, 03 July 2018

13. Typical Exam Questions
Higher Mathematics
Linear Relationships
Unit 1 Outcome 1
The Equation of the straight line
y – b = m (x - a)
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:
Tuesday, 03 July 2018