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Linear equations• Gradient of lines
• Graphing linear equations
• Finding the equation of a line
• Intersection of lines - simultaneous equations
• Simultaneous equations - elimination method
• Perpendicular lines
• Distance between points
• Midpoint of a line
VCE Maths Methods - Unit 1 - Linear equations
y
x
Gradient of lines
The gradient is the measure of how far up a line rises, as it it runs across.
gradient = rise
run=
y2 ! y1
x2 !x1
Rise = 4
Run = 8
This is a gradient of 4/8:this is the same as 1/2.
(2,2)
(10,6)
gradient = 6!2
10!2=
48=
12
Positive gradient Negative gradient
VCE Maths Methods - Unit 1 - Linear equations
Graphing equations - gradient form
Linear equations are defined by a gradient and y-intercept
y =mx+c
m = gradient
c = y intercept
y
x
(4,3)
Rise = 1Run = 4
y = x
4+2
(0,2)
VCE Maths Methods - Unit 1 - Linear equations
Graphing equations - intercept form
• Linear equations can also be wri!en in an intercept form: ax + by + c = 0
• The gradient form (y = mx + c) can be re-arranged.
y = x
2!4
0= x
2! y !4 0= x!2y !8
x intercept: y = 0 y intercept: x = 0
0= x!88= xx =8
0=!2y !88=!2y
y =!82=!4
y
x
(0,-4)
(8,0)
VCE Maths Methods - Unit 1 - Linear equations
Finding the equation of a line
• To find the equation of a line, a point and a gradient are needed.
• If two points are given, the gradient must be found first.
• The rule y - y1 = m(x - x1) is used to find the linear equation.
y intercept: x = 0
y
x
(1,2)
(9,-2)
Gradient:
m =
!2!29!1
=!48=!
12
Equation:
y ! y1=m(x!x1)
y !2=!1
2(x!1)
y !2=!1
2x+1
2
y =!1
2x+1
2+2
y =!1
2x+1
2+
42
y =!12
x+52
VCE Maths Methods - Unit 1 - Linear equations
Intersection of lines - simultaneous equations
• The intersection of two lines can be found by solving simultaneous equations.
• At the point of intersection, both lines have the same x & y values.
• The method of substitution can be used to find where one equation is equal to another.
• eg y = 2x - 4 and y = x - 1
2x!4= x!1
2x!x =!1+4
x =3
To find the y value - substitute this x value into either equation.
y =2(3)!4
y =2
VCE Maths Methods - Unit 1 - Linear equations
Intersection of lines - simultaneous equations
y
x
y = x -1
y = 2x -4
(3,2)
VCE Maths Methods - Unit 1 - Linear equations
Simultaneous equations - elimination method
• If the equations are given in intercept form, it is easier to use the elimination method to solve.
• Both equations should be lined up together & one variable eliminated by adding or subtracting the equations.
• eg 7x - 11y = -13 and x + y = 11
7x!11y =!13 x+ y =11
7x!11y =!13
7x+7 y =77Multiply by 7 to get 7x
in both equations
7x!7x!11y !7 y =!13!77 Subtract bo!om equationfrom the top one to cancel x
!18 y =!90
y =5 Simplify & solve
x+5=11 Find x by substituting into either equation x =6
VCE Maths Methods - Unit 1 - Linear equations
Perpendicular lines
Two lines are perpendicular if they cross at a 90° angle.
2
8
This is a gradient of 2/8:this is the same as 1/4.
8
2
This is a gradient of -8/2:this is the same as -4.
Two lines are perpendicular if their gradients multiply to -1.
m1!m2 ="1
VCE Maths Methods - Unit 1 - Linear equations
y
x
(2,-3)
(8,0)
d
Distance between points
The distance between two points can be found using Pythagoras’ theorem:
3
6
d2=62
+32
d2=36+9
d2=45
d = 45 =6.7
d = (x2 !x1)2+( y2 ! y1)2
VCE Maths Methods - Unit 1 - Linear equations
Midpoint of a segment
The midpoint of a straight line segment is at the middle of the x & y values.
y
x
(2,-2)
(10,4)
Midpoint
x value:
xm =
x1+x2
2
xm =
2+102
=122=6
y value:
y m =
y1+ y2
2
y m =
!2+42
=22=1
(6,1)