1: straight lines and gradients “teach a level maths” vol. 1: as core modules
TRANSCRIPT
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1: Straight Lines and 1: Straight Lines and GradientsGradients
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
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4
2
22
4 my
Finding the Gradient
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( 2, 3 )
( 0, 1)
3 ( 1) 4
2 0 2
02
)1(3m
2
4m 2m
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12
12
xx
yym
12 xx
12 yy
)3,2( ),( 22 yx
),( 11 yx )1,0(
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),( 22 yx
The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx
is
e.g. Find the gradient of the straight line joining the points and)1,0( )3,2(
22
4 mm
)1(
12
12
xx
yym
m 2 03
To use this formula, we don’t need a diagram!
),( 11 yxSolution:
12
12
xx
yym
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Equation of a Straight Line
The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx
is12
12
xx
yym
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Equation of a Straight Line
Find the gradient of the line joining the points A(3,–2) and B(–5,6)
12
12
xx
yym
x2 = –5 y2 = 6x1= 3 y1= –2
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Equation of a Straight Line
Find the gradient of the line joining the points A(3,–2) and B(–5,6)
12
12
xx
yym
x2 = –5 y2 = 6
35
)2(6
m 35
26
8
8
1
x1 = 3 y1= –2
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Equation of a Straight Line
c is the point where the line meets the y-axis, the y-intercept
and y-intercept, c = 2
1e.g. has gradient m = 12 xy
cmxy • The equation of a straight line ism is the gradient of the line
gradient = 2
x
12 xy
intercept on y-axis
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Equation of a Straight Line
gradient = 2
x
12 xy
intercept on y-axis
( 4, 7 )x
• The coordinates of any point lying on the line satisfy the equation of the line
showing that the point ( 4,7 ) lies on the line.
71)4(2 yye.g. Substituting x = 4 in gives12 xy
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Equation of a Straight Line Finding the equation of a straight line when
we know
e.g.Find the equation of the line with gradient passing through the point
)3,1( 2
• its gradient, m and • the coordinates of a point on the line
(x1,y1).
Solution:
y m and x3, 2 1
So, 52 xy
y m x c c3 2 1
y x2 5
Using , m is given, so we can find c bysubstituting for y, m and x.
y m x c
(-1, 3)
52 xy
x
c3 2 c3 2
Add 2 to both sidesC = 5
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Equation of a Straight Line
Solution: First find the gradient
We could use the 2nd point,(-1, 3) instead of (2, -3)
Using the formula when we are given two points on the line
cmxy
e.g. Find the equation of the line through the points )3,1()3,2( and
12
12
xx
yym
12 xy
2)1(
)3(3
m3
6
m 2 m
Now use with
cmxy 3 and2 yx
c 223
1 c
),( 11 yx 2 2( , )x y
-3 = -4 + c
Add 4 to both sides
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Equation of a Straight Line
SUMMARY
Equation of a straight line
Gradient of a straight line
12
12
xx
yym
y mx c
where and are points on the line
),( 11 yx ),( 22 yx
where m is the gradient and c is the intercept on the y-axis
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Equation of a Straight Line
2. Find the equation of the line through the points )4,1()2,1( and
Exercise1. Find the equation of the line with gradient 2
which passes through the point . )1,4( Solution: 92 xySo,
Solution:
y ym
x x2 1
2 1
m
4 23
1 ( 1)
13 xySo,
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Equation of a Straight Line
We sometimes rearrange the equation of a straight line so that zero is on the right-hand side ( r.h.s. )
We must take care with the equation in this form.
e.g. can be written as
12 xy 012 yx
e.g. Find the gradient of the line with equation
0734 yxSolution: Rearranging to the form :
cmxy
0734 yx 743 xy
3
7
3
4
x
y
)( cmxy
so the gradient is 3
4