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)