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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
Perpendicular Lines
The gradient of the straight line joining the pointsand),( 11 yx ),( 22 yx
is12
12
xx
yym
Perpendicular Lines
-1
-2
-3
-4
-5
1
2
3
4
5
0X->
|̂Y
Gradient of black line =
23
Gradient of red line =
32
3
2
2
–3
Perpendicular Lines Flip the gradient and change the sign
Perpendicular Lines
Perpendicular Lines Flip the gradient and change the sign to find the perpendicular gradient
1
mie gradient perpendicular to m is –
So gradient perpendicular to m = 3 is m = –
So gradient perpendicular to m = is m = –
If two lines are perpendicular then m1m2 = –1
Perpendicular Lines
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
Perpendicular Lines
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: y y m x x1 1( )
on the line
)4,1( y x1
4 ( 1)2
2
9
2
1 xy
92 xy 092 yxor
Perpendicular LinesExercise
Solution:
12 xy
on line y x(2, 3) 3 2( 2) 12 xySo,
Solution: 2
321032 xyxy
21
1 m
012xy
12
1
mm
on line y x(1,2) 2 1( 2)
xy 2So,
1. Find the equation of the line parallel to the line
which passes through the point .
)3,2( 012 xy2 m
Parallel line is has a gradient of –2
2. Find the equation of the line through the point (1, 2), perpendicular to the line 032 xy
22 m
Perpendicular Lines
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
