A straight line is fixed by two pieces of information, and its intercepts are two of the easiest to use. The x-intercept is the point where the line meets the x-axis, so it has the form (a, 0). The y-intercept is where it meets the y-axis, so it has the form (0, b).
This skill appears at the start of straight lines and linearisation and returns whenever a question says “the line crosses the axes at” or asks for the area of a triangle formed with the axes.
How do two intercepts give you an equation?
Write the two intercepts as points, then use them exactly like any other two points. The gradient is (change in y) ÷ (change in x), and the y-intercept is already the constant in y = mx + c.
There is also a direct form. A line with x-intercept a and y-intercept b can be written x/a + y/b = 1, and its gradient is −b/a. Both routes give the same line, so use whichever you can do without slips.
Step by step
- Write the two points, for example (a, 0) and (0, b).
- Find the gradient using the same order for both coordinates: m = (y₂ − y₁) ÷ (x₂ − x₁).
- Read the y-intercept straight from the point (0, b), so c = b.
- Write y = mx + c, then check both points satisfy it.
If you are given one intercept and a gradient instead, use y − y₁ = m(x − x₁) with the intercept point, then rearrange.
Worked example
A line meets the x-axis at (6, 0) and the y-axis at (0, −4). Find its equation and the area of the triangle it forms with the axes.
Step 1, points: (6, 0) and (0, −4).
Step 2, gradient: m = (−4 − 0) ÷ (0 − 6) = −4 ÷ −6 = 2/3.
Step 3, y-intercept: c = −4.
Step 4, equation: y = (2/3)x − 4.
Check: at x = 6, y = 4 − 4 = 0, so (6, 0) is on the line.
Area: the triangle has base 6 and height 4 (lengths, so positive). Area = ½ × 6 × 4 = 12 square units.
The mistake to watch for
A common slip is to subtract the coordinates in different orders.
Mistaken working: m = (0 − (−4)) ÷ (0 − 6) = 4 ÷ (−6) = −2/3
The numerator used (6, 0) first, but the denominator used (0, −4) first.
A quick sketch exposes it. The line passes through (0, −4) below the axis and (6, 0) on the axis, so it rises to the right and the gradient must be positive. The correction is to label each point once as (x₁, y₁) and (x₂, y₂) and subtract in the same direction on top and bottom.
Check yourself
Try these on paper, then open each answer.
1. A line crosses the axes at (−3, 0) and (0, 5). Find its equation.
Show answer
m = (5 − 0) ÷ (0 − (−3)) = 5/3. The y-intercept is 5.
y = (5/3)x + 5
Check: at x = −3, y = −5 + 5 = 0.
2. A line has gradient 4 and passes through the x-intercept (2, 0). Find its equation and its y-intercept.
Show answer
y − 0 = 4(x − 2), so y = 4x − 8.
y = 4x − 8, y-intercept −8.
Check: at x = 2, y = 8 − 8 = 0.
3. The line 3x + 4y = 12 meets the axes at A and B. Find A, B and the area of the triangle OAB, where O is the origin.
Show answer
Put y = 0: 3x = 12, so x = 4 and A = (4, 0). Put x = 0: 4y = 12, so y = 3 and B = (0, 3).
Area = ½ × 4 × 3 = 6 square units.
Where this leads next
Next, see how perpendicular lines use the gradient you just found, or go straight to transforming data into straight-line form. The non-calculator working trainer is handy for keeping fractional gradients exact, and the mixed practice set tests the whole module.
Some students can find a gradient from two points but freeze when the points are hidden inside wording like “meets the axes at”. That is the kind of gap our teachers look for in online one-to-one Additional Mathematics tuition.