1 straight lines and gradients
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1: Straight Lines and 1: Straight Lines and GradientsGradients
© Christine Crisp
““Teach A Level Maths”Teach A Level Maths”
Vol. 1: AS Core Vol. 1: AS Core ModulesModules
Straight Lines and Gradients
Module C1
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Straight Lines and Gradients
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
Straight Lines and Gradients
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
Straight Lines and Gradients
Notice that to find c, the equation has been solved from right to left. This takes a bit of practice but reduces the chance of errors.
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.
Solution: 12,3 xmy and
So, 52 xy
ccmxy )1(23c 5
Using , m is given, so we can find c bysubstituting for y, m and x.
cmxy
(-1, 3)
52 xy
x
Straight Lines and Gradients
If we don’t know the gradient, we have to find it using two points on the line.
We develop the formula by reminding ourselves about the meaning of a gradient.
To do this, we can use a formula.
Straight Lines and Gradients
4
2
e.g.
22
4 mm
Straight Lines and Gradients
( 2, 3 )
( 0, 1)
3 ( 1) 4
2 0 2
02
)1(3m
2
4m 2m
Straight Lines and Gradients
12
12
xx
yym
12 xx
12 yy
)3,2( ),( 22 yx
),( 11 yx )1,0(
Straight Lines and Gradients
12
12
xx
yym
m
),( 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(2 0
3
To use this formula, we don’t need a diagram!
),( 11 yxSolution:
12
12
xx
yym
Straight Lines and Gradients
To find the equation of a straight line given 2 points on the line.
Solution: First find the gradient:
e.g. Find the equation of the line through the points
)3,1()3,2( and
12
12
xx
yym
Now
cmxy on the line:
)3,2( c )2(23c 1
Equation of line is
12 xy
3
6
m
2)1(
)3(3
m
2 m
cxy 2
Straight Lines and Gradients
SUMMARY
Equation of a straight line
Gradient of a straight line
12
12
xx
yym
cmxy
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
Straight Lines and Gradients
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:
cxy 2c )4(21)1,4( line on c 9
92 xySo,
Solution:
12
12
xx
yym
3)1(1
24
m
cmxy
cmxy cxy 3c )1(32)2,1( line on c 1
13 xySo,
Straight Lines and Gradients
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
Straight Lines and Gradients
Parallel and Perpendicular Lines
They are parallel if 12 mm
They are perpendicular if 1
21
mm
If 2 lines have gradients and , then:1m 2m
Straight Lines and Gradients
e.g. 1 Find the equation of the line parallel towhich passes through the point )3,1(
12 xy
Solution: The given line has gradient 2. Let 21 mFor parallel lines,
22 m12 mm
is the equation of any line parallel to
Using cmxy cxy 2
12 xyon the line
)3,1( c )1(23
c 1
12 xy
Straight Lines and Gradients
We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:
e.g.Find the equation of the line perpendicular to passing through the point . 12 xy )4,1(
Solution: The given line has gradient 2. Let
21 m
Perpendicular lines:2
12 m
12
1
mm
Equation of a straight line: cmxy on the line
)4,1( c )1(2
14 c
2
9
2
9
2
1 xy
92 xy 092 yxor
Straight Lines and Gradients
• If the gradient isn’t given, find the gradient using
Method of finding the equation of a straight line:
• Substitute for y, m and x in intoto find c.
cmxy
either parallel lines: 12 mm
or 2 points on the line:
12
12
xx
yym
or perpendicular lines:1
21
mm
SUMMARY
Straight Lines and GradientsExercise
Solution:
12 xy
c )2(23)3,2( line on c 112 xySo,
Solution: 2
321032 xyxy
21
1 m
012xy
12
1
mm cxy 2
c 22)2,1( line on c 0xy 2So,
1. Find the equation of the line parallel to the line
which passes through the point .
)3,2( 012 xy2 mParallel line
iscxy 2
2. Find the equation of the line through the point (1, 2), perpendicular to the line 032 xy
22 m So,
Straight Lines and Gradients
A Second Formula for a Straight Line ( optional )Let ( x, y ) be any point on the line
1xx
1yy
1
1
xx
yym )( 11 xxmyy
Let be a fixed point on the line
),( 11 yx
),( yxx
),( 11 yx x
Straight Lines and Gradients
Solution: First find the gradient
We could use the 2nd point,(-1, 3) instead of (2, -3)
To use the formula we need to be given
either: one point on the line and the gradient
or: two points on the line
)( 11 xxmyy
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 )( 11 xxmyy 32 11 yx and
)2)(2()3( xy423 xy
Straight Lines and Gradients
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Straight Lines and Gradients
They are parallel if 12 mm
They are perpendicular if 1
2
1
mm
If 2 lines have gradients and , then:1m 2m
Equation of a straight line
Gradient of a straight line
12
12
xx
yym
cmxy
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
SUMMARY
Straight Lines and Gradients
Solution: First find the gradient:
e.g. Find the equation of the line through the points
)3,1()3,2( and
12
12
xx
yym
2)1(
)3(3
m
3
6
m 2 m
Now
cmxy cxy )(2on the line:
)3,2( c )2(23c 1
Equation of line is
12 xy
Straight Lines and Gradients
We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:
e.g.Find the equation of the line perpendicular to passing through the point . 12 xy )4,1(
Solution: The given line has gradient 2. Let
21 m
Perpendicular lines:2
12 m
12
1
mm
Equation of a straight line: cmxy on the line
)4,1( c )1(2
14 c
2
9
2
9
2
1 xy
92 xy 092 yxor