Solving Problems Using Linear Equations
Worked Example 6:
Mira is twice Sanjin’s age. How old are Mira and Sanjin now if in 10 years the sum of their ages will be 89. Use s to represent Sanjin’s age.
Sanjin’s Age: s Mira’s present age: 2s
In 10 years:
Sanjin = s + 10 Mira’s age = 2s + 10
(s+10) + (2s+10) = 89
3s + 20 = 89
3s = 69
s = 23
2s = 2 x 23 = 46
Sanjin is 23 years old and Mira is 46 years old. In 10 years they will be 33 and 56 old respectively; 33 + 56= 89 
Distance and Midpoint
Distance Formula 
Midpoint Formula 
The distance, d, between any two Cartesian coordinates (x1, y1) and (x2, y2) is: 
The midpoint M(x,y) between 2 points (x1, y1) and (x2, y2) is given by: 
d = 
M(x,y) = 


Gradient
The gradient of a line can be found by the ratio of the vertical rise to the horizontal run between any two points on the line.
Gradient AB = (vertical rise)/(horizontal run)
The letter m is used to represent the gradient. y=mx+b
m = (rise)/(run)
= (change in yvalue)/(change in xvalue)
= (y2y1)/(x2x1) 
Linear Graphs: Gradient and YIntercept
The gradientintercept form:
y=mx+b, where m is the gradient and b is the yintercept.
The general form:
ax+by+c=0, where a, b and c are constants and a ≠ 0, b ≠ 0.
In this equation, the gradient m = (a)/(b) and the yintercept = (c)/(b)can be found by rearranging the equation to make y the subject.
Worked Example 15
(a) y = 3x+2
gradient = 3
yintercept: (0,2)
(b) y = x1
= 1x+(1)
gradient = 1
yintercept: (0,1)
(c) 2y = 5x  4
y = (5)/(2)x  2
gradient = (5)/(2)
yintercept: (0,2)
(d) y+2x =6
y = 62x
y = 2x+6
gradient: 2
yintercept: (0,6) 


Sketching Linear Graphs
Graphing using the yintercept and gradient
There is no need to plot a number of points to graph linear relationships. Because only 2 points are needed to define a line, use the yintercept as one point on the graph and the gradient, m to locate the second point. Once 2 points are known, a straight line can be drawn through them.
Sketching other Linear Graphs, y=mx+b, b ≠ 0
If the yintercept is not the origin, then first identify the yintercept from the equation before using the gradient to find the second point.
Using the equation y=mx+b and substituting x = 0, you get y = b.
x = 0 is the x=coordinate of a point on the yaxis, so b is the yvalue at the point at the point at which the graph crosses the yaxis. This is the yintercept
Worked Example 17
Use the yintercept and the gradient to sketch the graph of y = 2x+1
yintercept: (0,1)
Gradient = 2
= (2)/(1) = (rise)/(run) 
Parallel and Perpendicular Lines
Worked Example 19
Find the gradient of the line parallel to each of the following
(a) y = 2x5
m = 2
Any parallel line will have a gradient of 2.
(b) 2x + 3y = 6
2x2x+3y = 6  2x
(3y)/(3) =  (2)/(3)x + (6)/(3)
y =  (2)/(3)x + 2
m1 =  (2)/(3)
Any parallel line will have a gradient of  (2)/(3).
Worked Example 20
Find the gradient of a line perpendicular to each of the following
(a) y = 4x + 3
m1 = 4
m2 =  (1)/(m1)
=  (1)/(4)
(b) 3x5y = 10
3x3x5y = 103x
5y = 3x + 10
(5y)/(5) = (3x)/(5) + (10)/(5)
y = (3)/(5)x  2 m1= (3)/(5)
m2 =  (1)/(m1) = (5)/(3) 

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