coordinate geometry © christine crisp. straight lines and gradients c is the point where the line...
TRANSCRIPT
Coordinate geometryCoordinate geometry
© Christine Crisp
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
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
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 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
Finding the Mid-Point of AB
The mid-point is the average of the end points:
M X
)2,2( A
)3,1(B
2
1
2
12
Mx
2
1
2
32
My
Finding the Mid-Point of AB
A formula for the Mid-Point of AB
M X
or
)2,2( A ),( 11 yxA
)3,1(B ),( 22 yxB
,2
21 xxx M
221 yy
y M
The mid-point is the average of the end points:
2
,2
2121 yyxxM is
Finding the Mid-Point of AB
Solution:
Exercise
Find the mid-point, M of the line joining to
)3,4( )2,1(
2
,2
2121 yyxxM is
2
23
2
)1(4,isM
2
1,
2
3
Length of the line joining A to B
Using Pythagoras’ theorem:
)2,2( A
)3,1(B
3)2(1
5)2(3
3453 22 AB
Length of the line joining A to B
A formula for the length of the line joining A to B
Using Pythagoras’ theorem:
)2,2( A ),( 11 yxA
)3,1(B ),( 22 yxB
12 xx
12 yy
212
212 )()( yyxxAB
Straight Lines and Gradients
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