straight lines and gradients objectives: to find linear equations from minimum information. to use...
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Straight Lines and Gradients
Objectives:To find linear equations from
minimum information.To use linear equations in
any form to find the gradient and y-axis intercept.
Straight Lines and Gradients
Objectives:To find linear equations from
minimum information.To use linear equations in
any form to find the gradient and y-axis intercept.
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
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
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
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
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
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
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
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
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
• 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
Exercise
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 m
Parallel line is
cxy 2
2. Find the equation of the line through the point (1, 2), perpendicular to the line 032 xy
22 m So,
A Second Formula for a Straight Line ( really useful )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
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
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
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