A line passes through:
(2,5)
and:
(6,13).
Find its equation.
To reconstruct a straight line, we need enough information to determine both its rate of change and its position.
Method 1: two points
First find the gradient:
m=(13−5)/(6−2)=8/4=2.
Now use:
y=mx+c.
Substitute one point, say (2,5):
5=2(2)+c.
c=1.
Equation:
y=2x+1.
The quick answer
- Find the gradient if it is not already given.
- Use y=mx+c or point–slope form.
- Substitute a known point.
- Solve for the remaining constant.
- Check the equation using another known point.
Method 2: point plus gradient
A line has gradient 3 and passes through (4,10).
Write:
y=3x+c.
Substitute:
10=3(4)+c.
c=−2.
So:
y=3x−2.
Point–slope form
If a line has gradient m and passes through (x₁,y₁), then:
y−y₁=m(x−x₁).
Using m=3 and point (4,10):
y−10=3(x−4).
Expand:
y−10=3x−12.
Therefore:
y=3x−2.
Point–slope form packages the gradient and one known point directly; y=mx+c makes gradient and intercept easier to read.
Method 3: from a graph
Suppose a graph shows:
- y-intercept=−3;
- another clear point=(2,1).
Gradient:
(1−(−3))/(2−0)=4/2=2.
So:
y=2x−3.
Use clear points, not visually convenient points
When reading a graph:
- choose points the line clearly passes through;
- read actual axis values;
- prefer points far apart;
- do not infer coordinates from the apparent angle of the line.
Graph-reading error can distort both gradient and intercept.
Finding the line from an x-intercept and y-intercept
Suppose the line crosses:
- x-axis at (4,0);
- y-axis at (0,6).
Gradient:
(0−6)/(4−0)=−6/4=−3/2.
Since c=6:
y=−(3/2)x+6.
Horizontal lines
If a line passes through (2,5) and (8,5), its gradient is 0.
Every point has y=5.
Equation:
y=5.
Vertical lines
If a line passes through (3,1) and (3,7), x is always 3.
Equation:
x=3.
This cannot be written in y=mx+c form with a finite gradient.
Worked example with a fractional gradient
Line through (2,1) and (8,4).
Gradient:
(4−1)/(8−2)=3/6=1/2.
Use point (2,1):
1=(1/2)(2)+c.
c=0.
Equation:
y=x/2.
This is also direct proportion.
Worked example with negative gradient
Line through (−1,5) and (3,−3).
m=(−3−5)/(3−(−1))=−8/4=−2.
Using (−1,5):
5=−2(−1)+c.
c=3.
Equation:
y=−2x+3.
Why one point is not enough by itself
Infinitely many lines pass through one point.
For example, through (0,2):
- y=x+2;
- y=2x+2;
- y=−5x+2
all pass through the point.
We need additional information such as gradient or a second point.
Why two distinct points are enough
Two distinct points determine one straight line.
The points determine the gradient.
Once the gradient is fixed, one point determines the vertical position.
Checking the equation with both points
For y=2x+1 and points (2,5), (6,13):
- x=2 → y=5;
- x=6 → y=13.
Both points satisfy the equation.
A proposed line equation should reproduce every piece of evidence used to construct it.
Equation from context
A machine costs $20 to start and $6 per hour to run.
Let x be hours and y total cost.
Gradient=6 dollars/hour.
Intercept=20 dollars.
Equation:
y=6x+20.
Sometimes the line can be written directly from its meaning without first calculating from points.
Equation from a table
Suppose:
- x=1 → y=7;
- x=3 → y=15;
- x=5 → y=23.
Gradient:
(15−7)/(3−1)=8/2=4.
Use x=1,y=7:
7=4+c.
c=3.
Equation:
y=4x+3.
Common misconception 1: use y/x for gradient
Use change in y divided by change in x unless the line is known to pass through the origin.
Common misconception 2: substitute x and y into the wrong positions
In y=mx+c, y is the vertical coordinate and x is the horizontal coordinate.
Common misconception 3: one point determines the line
One point needs additional information such as gradient.
Common misconception 4: every line belongs in y=mx+c form
Vertical lines are written x=constant.
A line-equation diagnostic ladder
- Can the learner find gradient from two points?
- Can the learner use y=mx+c with a known point?
- Can the learner use point–slope form?
- Can the learner find an equation from a graph?
- Can the learner use intercepts as points?
- Can the learner handle fractional and negative gradients?
- Can the learner identify horizontal and vertical lines?
- Can the learner construct a line from a table or context?
- Can the learner verify the equation using a second point?
How this fits Secondary Mathematics
Finding straight-line equations connects coordinates, gradient, algebraic substitution, graphs and modelling. It also prepares learners for parallel and perpendicular lines, simultaneous equations and coordinate geometry.
The deeper lesson: reconstruct the rule from evidence
A graph, a pair of points and a rate of change may look like different kinds of information.
For a straight line, they are different views of one underlying relationship.
Finding a line equation means recovering the hidden rule that produced the visible evidence: determine how the line changes, anchor it to a known point, and verify that the resulting equation reproduces the geometry.