Equations of linear graphs
Section: Coordinate geometry | Syllabus: Cambridge IGCSE Mathematics (0580)
Interpreting y = mx + c
Every non-vertical straight line can be written in the form y=mx+c, where m is the gradient and c is the y-intercept. Once a line's equation is in this form, both features can be read off directly.
- In y=mx+c, m is the gradient (the coefficient of x) and c is the y-intercept (where the line crosses the y-axis)
- If an equation is not already in this form, rearrange it first so that y is the subject
- Two lines with the same gradient m are parallel
Worked Example: Reading the Gradient and Intercept
- Question: A line has equation 2y=6x-10. Write down its gradient and y-intercept.
- Step 1: Rearrange so y is the subject - divide every term by 2: y=3x-5
- Step 2: Compare with y = mx + c: m = 3, c = -5
- Answer: gradient = 3, y-intercept = -5
Common Mistakes
MistakeReading off the gradient and intercept before rearranging the equation into the form y = mx + c
Fixalways check that y has a coefficient of exactly 1 (is fully the subject) before reading off m and c
MistakeDividing only one term by the coefficient of y, rather than every term in the equation
Fixwhatever is done to rearrange the equation must be done to every single term, not just the ones already containing x
Finding the Equation of a Line from a Gradient and a Point
If the gradient of a line and one point on it are known, the full equation y = mx + c can be found by substituting both into the general form.
- Substitute the known gradient for m in y = mx + c
- Substitute the coordinates of the known point for x and y, then solve for c
- Write the final equation using the values found for m and c
Worked Example: Finding an Equation from a Gradient and a Point
- Question: A line has gradient 4 and passes through the point (3, 5). Find the equation of the line in the form y = mx + c.
- Step 1: Substitute m = 4: y=4x+c
- Step 2: Substitute the point (3, 5): 5=4(3)+c
- Step 3: Solve: 5=12+c c=-7
- Answer: y = 4x - 7
Common Mistakes
MistakeSubstituting the point's coordinates the wrong way round, e.g. putting the y-value where x should go
Fixin the point (a,b), a is always the x-coordinate and b is always the y-coordinate - substitute accordingly into x and y
MistakeForgetting to solve for c after substituting, and leaving c undetermined in the final equation
Fixafter substituting, rearrange the resulting equation to find the numerical value of c before writing the final answer
Finding the Equation of a Line from Two Points
When only two points on a line are known, the gradient must be calculated first, before the equation can be found using the same method as before.
- First calculate the gradient using the two points: m=(y_2-y_1/x_2-x_1)
- Then substitute the gradient and either one of the two points into y = mx + c to find c
- Either of the two given points can be used in the second step - both will give the same value of c
Worked Example: Finding an Equation from Two Points
- Question: A line passes through the points (1, 4) and (4, 13). Find the equation of the line in the form y = mx + c.
- Step 1: Find the gradient: m=(13-4/4-1)=(9/3)=3
- Step 2: Substitute m = 3 and the point (1, 4): 4=3(1)+c
- Step 3: Solve: 4=3+c c=1
- Answer: y = 3x + 1
Common Mistakes
MistakeTrying to find c before calculating the gradient, leaving two unknowns in the equation at once
Fixalways find the gradient first - it must be known before c can be found
MistakeUsing values from the two points inconsistently, e.g. the x-coordinate from one point and the y-coordinate from the other
Fixalways substitute a complete matching pair (x, y) from the same point when solving for c
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