higher linear relationships lesson 7
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Higher Linear Relationships Lesson 7. Linear Relationships. 6. 0. Which of the following points does not lie on the line 2 y + 5x – 4 = 0?. (-0.8, 0) (-1,0.5) (0,2) (2,3) I don’t know. Linear Relationships. 6. 0. - PowerPoint PPT PresentationTRANSCRIPT
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
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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
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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
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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
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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
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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
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The Equation of the straight liney – b = m (x - a)