Linear Relationships Cheat Sheet

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

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)
= (y2-y1­)/(­x2-x1)

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
y-inte­rcept: (0,2)

(b) y = x-1
= 1x+(-1)
gradient = 1
y-inte­rcept: (0,-1)

(c) 2y = 5x - 4
y = (5)/(2)x - 2
gradient = (5)/(2)
y-inte­rcept: (0,-2)

(d) y+2x =6
y = 6-2x
y = -2x+6
gradient: -2
y-inte­rcept: (0,6)

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
y-inte­rcept: (0,1)
Gradient = 2
= (2)/(1) = (rise)­/(run)

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)