A straight-line graph is a compressed statement about constant rate of change. Its gradient tells us how much one quantity changes when another changes by one unit. Its intercept tells us the starting value or fixed offset. Together they turn a visual pattern into an equation.
This thirtieth Secondary 4 Mathematics Learning Guide develops straight-line graphs, gradient, intercepts and rate of change as one connected representation system. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map.
The guide focuses on coordinate geometry, equations of straight lines, interpretation of gradient and intercept, parallel and perpendicular relationships, graph modelling and graphical verification.
Gradient measures change in y per unit change in x
For two points (x₁,y₁) and (x₂,y₂), the gradient m is:
m=(y₂−y₁)/(x₂−x₁).
A positive gradient rises from left to right. A negative gradient falls. Zero gradient gives a horizontal line. A vertical line has undefined gradient because x₂−x₁=0.
Worked Example 1 | Find gradient from two points
Find the gradient of the line through A(2,5) and B(8,17).
m=(17−5)/(8−2)=12/6=2.
The line rises 2 units in y for every 1 unit increase in x.
The equation y=mx+c separates rate and starting value
In y=mx+c, m is gradient and c is the y-intercept, the value of y when x=0.
For y=3x−4, the gradient is 3 and the y-intercept is −4. The line therefore passes through (0,−4) and rises 3 units for every 1 unit moved right.
Worked Example 2 | Form an equation from gradient and intercept
A line has gradient −2 and y-intercept 7. Its equation is:
y=−2x+7.
At x=0, y=7. At x=1, y=5. The change confirms the gradient −2.
Find a line equation from one point and a gradient
If m is known and a point lies on the line, substitute the point into y=mx+c to find c.
Worked Example 3 | Point and gradient
Find the equation of the line with gradient 4 through (3,5).
Use y=4x+c. Substitute (3,5):
5=12+c, so c=−7.
y=4x−7.
Find a line equation from two points
First calculate the gradient, then substitute either point into y=mx+c.
Worked Example 4 | Two-point equation
Find the equation through (1,4) and (5,12).
Gradient=(12−4)/(5−1)=8/4=2.
Use y=2x+c. Substitute (1,4): 4=2+c, so c=2.
y=2x+2.
Check with the second point: 2(5)+2=12.
The x-intercept solves y=0
The x-intercept is where the graph crosses the x-axis, so y=0. For y=2x−6:
0=2x−6, so x=3.
The x-intercept is therefore (3,0).
Worked Example 5 | Intercepts as quick plotting anchors
For 3x+2y=12, find both intercepts.
Set y=0: 3x=12, so x=4.
Set x=0: 2y=12, so y=6.
The line passes through (4,0) and (0,6).
Two distinct points are enough to determine a straight line.
Parallel lines have the same gradient
Distinct parallel lines rise or fall at the same rate, so their gradients are equal. Their intercepts differ.
For example, y=3x+1 and y=3x−5 are parallel.
Worked Example 6 | Equation of a parallel line
Find the equation of the line parallel to y=−3x+8 and passing through (2,1).
The gradient is −3. Use y=−3x+c. Substitute (2,1):
1=−6+c, so c=7.
y=−3x+7.
Perpendicular gradients multiply to −1
For non-vertical, non-horizontal perpendicular lines with gradients m₁ and m₂:
m₁m₂=−1.
So if one gradient is 2, a perpendicular gradient is −1/2.
Worked Example 7 | Equation of a perpendicular line
Find the equation of the line perpendicular to y=2x+3 and passing through (4,1).
Perpendicular gradient=−1/2. Use y=−1/2 x+c.
1=−2+c, so c=3.
y=−1/2 x+3.
Gradient is rate of change in context
If y represents cost and x represents quantity, the gradient may represent cost per unit. If y is distance and x is time, gradient represents speed. If y is temperature and x is time, gradient represents temperature change per unit time.
The numerical gradient is incomplete until its units and meaning are stated.
Worked Example 8 | Cost model
A delivery service charges according to C=4.5d+8, where C is cost in dollars and d is distance in km.
- Gradient 4.5 means $4.50 per km.
- Intercept 8 means a $8 fixed charge when d=0.
For d=12 km, C=4.5(12)+8=$62.
Worked Example 9 | Distance-time graph
A straight segment on a distance-time graph goes from (2 h,70 km) to (5 h,250 km). Find the average speed represented by that segment.
Gradient=(250−70)/(5−2)=180/3=60 km/h.
The slope is not merely a geometric feature; it is the physical rate encoded by the axes.
Intersections solve simultaneous conditions
Where two lines intersect, both equations are true at the same coordinate pair. This is why graphical intersection and simultaneous-equation solution are the same mathematical event in different representations.
Worked Example 10 | Break-even point
Plan A costs 10+2x dollars. Plan B costs 25+x dollars. Find when the costs are equal.
Set 10+2x=25+x, so x=15.
At x=15, both cost 40.
Break-even point=(15,40).
Below 15 units, Plan A is cheaper. Above 15 units, Plan B is cheaper.
Horizontal and vertical lines
A horizontal line has equation y=k and gradient 0. A vertical line has equation x=k and undefined gradient.
A horizontal and vertical line are perpendicular even though the product-of-gradients rule cannot be applied because the vertical gradient is undefined.
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Subtracts x-values and y-values in inconsistent order | Gradient formula used mechanically | Keep point order consistent in numerator and denominator |
| Reads c as x-intercept | Intercept roles confused | Remember c is y when x=0 |
| Parallel line given reciprocal gradient | Parallel and perpendicular rules mixed | Parallel gradients are equal |
| Perpendicular gradient sign not changed | Negative reciprocal incomplete | Check product equals −1 |
| Gradient given without units in context | Graph geometry separated from meaning | Read both axis units |
| Vertical line written y=k | Coordinate condition reversed | Vertical means x stays constant |
Independent practice
- Find the gradient through (3,2) and (9,20).
- Find the equation of a line with gradient 5 and y-intercept −2.
- Find the equation through (2,7) and (6,15).
- Find both intercepts of 4x+2y=16.
- Find the equation through (1,3) parallel to y=−4x+9.
- Find the gradient of a line perpendicular to one with gradient 3/5.
Explained answers
1. (20−2)/(9−3)=18/6=3.
2. y=5x−2.
3. Gradient=(15−7)/(6−2)=2. Using (2,7): 7=4+c, so y=2x+3.
4. x-intercept (4,0); y-intercept (0,8).
5. Gradient −4. Use y=−4x+c and (1,3): c=7. y=−4x+7.
6. Negative reciprocal=−5/3.
Teaching sequence: graph, equation, meaning
Begin with coordinate changes and gradient as rise over run. Then connect gradient and intercept to y=mx+c. Move next to equations from points and to parallel/perpendicular conditions.
Finish with cost, distance and break-even models so every graphical feature must be interpreted in context.
Connect this guide to Simultaneous Linear Equations, Intersections and Modelling, Coordinate Geometry and Transformations as Representation, and Power, Reciprocal and Exponential Graphs.
Final thought
A straight line is a constant-rate machine. Gradient tells how fast the output changes. Intercept tells where the relationship starts. The equation and the graph are two views of the same structure.
Read the axes, read the rate, then translate the line back into the relationship it represents.
Return to the Secondary Mathematics Hub.
