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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
Equation of a Straight Line
The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx
is12
12
xx
yym
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
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
Equation of a Straight Line
Formula for a Straight Line
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
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 x1 13, 2 1
So, 52 xy
y y m x x y x1 1( ) 3 2( 1)
y x2 5
Using , m is given, so we can find c bysubstituting for y1, m and x1.
y y m x x1 1( )
(-1, 3)
52 xy
xy x3 2 2
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
)( 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
Equation of a Straight Line
SUMMARY
Equation of a straight line
Gradient of a straight line
12
12
xx
yym
y mx c or y y m x x1 1- ( )
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
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:
on line y x(4, 1) 1 2( 4) 92 xySo,
Solution:
y ym
x x2 1
2 1
m
4 23
1 ( 1)
y y m x x1 1( )
on line y x( 1,2 ) 2 3( 1) 13 xySo,
y y m x x1 1( )
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
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
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
cm xy 3 and2 yx
c 223
1 c
),( 11 yx 2 2( , )x y
-3 = -4 + c
Add 4 to both sides
Equation of a Straight Line
The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.
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