higher linear relationships lesson 7

Higher

Linear Relationships

Lesson 7

Which of the following points does not lie on the line 2y + 5x – 4 = 0?

1 2 3 4 5

20% 20% 20%20%20%1. (-0.8, 0)

2. (-1,0.5)

3. (0,2)

4. (2,3)

5. I don’t know

1 2 3 4 5 6

0

6

Linear Relationships

1 3y = 4x + 5 2 4y = 3x - 1 3 4y + 3x = 7 4 4x + 3y = 2

1 2 3 4 5

20% 20% 20%20%20%1. Lines 1 and 2 are perpendicular

2. Lines 1 and 4 are parallel

3. Lines 2 and 4 are perpendicular

4. Lines 2 and 3 are parallel

5. I don’t know

1 2 3 4 5 6

0

6

Linear Relationships

A straight line has equation 10y = 3x + 15. Which of the following is true?

1 2 3 4 5

20% 20% 20%20%20%1. The gradient is 0.3 and the y intercept is 1.5

2. The gradient is 3 and the y intercept is 15

3. The gradient is 15 and the y intercept is 3

4. The gradient is 1.5 and the y intercept is 0.3

5. I don’t know

1 2 3 4 5 6

0

6

Linear Relationships

Higher Mathematics Linear Relationships Unit 1 Outcome 1

Wednesday 19 April 2023 www.chmaths.wikispaces.com

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

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.

Find the equations of the median and altitude.

Hence find the co-ordinates of K.

Draw a diagram

Higher Mathematics

P(-3,6)

Q(5,12)

R(5,2)

K

Wednesday 19 April 2023 www.chmaths.wikispaces.com

Example 1

Unit 1 Outcome 1Linear Relationships

Intersecting Lines

UsingP = (-3, 6) R = (5, 2)(1,4).

Gradient of median QK = (12 - 4) (5-1) mQK = 2

y - b = m(x - a)

y - 4 = 2(x - 1)

y - 4 = 2x - 2

2x - y = -2

mQR = (12 - 2) / (5 - 5) = 10 / 0 = undefined mPK

PK is zero

y - b = m( x - a) y - 6 = 0(x + 3)

y - 6 = 0

y = 6

Q(5,12)

KP(-3,6)

R(5,2)

Higher Mathematics Linear Relationships Unit 1 Outcome 1Intersecting 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 Find the midpoint of PR

x-3 + 5 2

y6 + 2 2

Midpoint of PR , K, is (1,4)

To find the equation of the altitude PK

Find gradient QR

www.chmaths.wikispaces.com

To find the point of intersection of K

2x - y = -2 (A)

y = 6 (B)

Substitute y = 6 in the top equation

2x - 6 = -2

2x = 4

x = 2

Hence K is the point (2,6)

P(-3,6)

Q(5,12)

R(5,2)

K

Higher Mathematics Linear Relationships Unit 1 Outcome 1

Intersecting Lines

Wednesday 19 April 2023 www.chmaths.wikispaces.com

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.

Hence find the co-ordinates of the centre of the circle, C.

Diagram will be something like

E(-1,1)

F(3,3)

G(6,2)

C

Higher Mathematics Linear Relationships Unit 1 Outcome 1

Midpoint of EF is A (1,2)

mEF = (3-1)/(3+1) = 1/2

y - b = m(x - a)

y - 2 = -2( x - 1)

y - 2 = -2x + 2

2x + y = 4

Find Midpoint of FG

mFG = (2-3)/(6-3) = -1/3

Perpendicular m = 3

y - b = m(x – a)

y - 2.5 = 3( x - 4.5)

y - 2.5 = 3x - 13.5

3x - y = 11

E(-1,1)

F(3,3)

G(6,2)

C

Perpendicular m = -2

Higher Mathematics Linear Relationships Unit 1 Outcome 1

Find the equations of the perpendicular bisectors of EF and FG.

Find Midpoint of EF UsingE = (-1, 1) F = (3, 3)

x-1 + 3 2

y1 + 3 2

A

Midpoint of FG is B (4.5, 2.5)

UsingF = (3, 3) G = (6, 2)

x3 + 6 2

y3 + 2 2

B

A B

Finding where the bisectors meet gives us the centre of the circle

Solving

2x + y = 4 (A)

3x - y = 11 (B)

5x = 15

x = 3

Substituting x = 3 into A

6 + y = 4

y = -2

Hence centre of circle at (3,-2)

E(-1,1)

F(3,3)

G(6,2)

C

2x + y = 4

Linear Relationships Unit 1 Outcome 1

Wednesday 19 April 2023

Unit 1 Outcome 1Higher Mathematics

www.chmaths.wikispaces.com

Classwork

Page 11 Exercise 6Complete

Homework

Page xx Exercise AQuestion xxxx

Linear Relationships

The Equation of the straight lineIntersecting Straight Lines using

simultaneous equations

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 Questions

Linear RelationshipsHigher Mathematics Unit 1 Outcome 1

Wednesday 19 April 2023 www.chmaths.wikispaces.com

The Equation of the straight liney – b = m (x - a)