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
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:
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 y-value)/(change in x-value)
Linear Graphs: Gradient and Y-Intercept
The gradient-intercept form:
y=mx+b, where m is the gradient and b is the y-intercept.
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 y-intercept = (-c)/(b)can be found by rearranging the equation to make y the subject.
Worked Example 15
(a) y = -3x+2
gradient = -3
(b) y = x-1
gradient = 1
(c) 2y = 5x - 4
y = (5)/(2)x - 2
gradient = (5)/(2)
(d) y+2x =6
y = 6-2x
y = -2x+6
Sketching Linear Graphs
Graphing using the y-intercept 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 y-intercept 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 y-intercept is not the origin, then first identify the y-intercept 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 y-axis, so b is the y-value at the point at the point at which the graph crosses the y-axis. This is the y-intercept
Worked Example 17
Use the y-intercept and the gradient to sketch the graph of y = 2x+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 = 2x-5
m = 2
Any parallel line will have a gradient of 2.
(b) 2x + 3y = 6
2x-2x+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) 3x-5y = 10
3x-3x-5y = 10-3x
-5y = -3x + 10
(-5y)/(-5) = (-3x)/(-5) + (10)/(-5)
y = (3)/(5)x - 2 m1= (3)/(5)
m2 = - (1)/(m1) = -(5)/(3)