x-y Intercepts

Every straight line crosses the two axes at special points. Set \(x = 0\) to find where it meets the \(y\)-axis, and \(y = 0\) to find where it meets the \(x\)-axis. Two quick substitutions, and you can sketch the whole line.

MYP 4Standard MathsCoordinate GeometryCriteria A · C · D~10 min read

What an intercept is

An intercept is simply a point where a line crosses an axis. Every non-vertical, non-horizontal straight line crosses both axes exactly once, so it has two of them.

The key idea
On the \(y\)-axis, every point has \(x = 0\). On the \(x\)-axis, every point has \(y = 0\). So to find an intercept, you set the other coordinate to zero and solve for what is left.

The y-intercept: set \(x = 0\)

To find where a line crosses the \(y\)-axis, put \(x = 0\) into the equation and work out \(y\). For example, in \(y = 4x + 5\), setting \(x = 0\) gives \(y = 5\), so the line crosses at \((0, 5)\). With an equation already in the form \(y = mx + c\), the \(y\)-intercept is just \(c\).

The x-intercept: set \(y = 0\)

To find where a line crosses the \(x\)-axis, put \(y = 0\) and solve for \(x\). In \(y = 4x + 5\), setting \(y = 0\) gives \(0 = 4x + 5\), so \(x = -\tfrac{5}{4}\) and the line crosses at \(\left(-\tfrac{5}{4}, 0\right)\).

Zero the opposite letter

The word "\(y\)-intercept" tells you where \(x = 0\), not \(y = 0\). It is the axis you land on that keeps its value, so set the other one to zero.

Sketching from both intercepts

Two points fix a straight line, so once you have both intercepts you can join them with a ruler and you are done. This works cleanly for equations written as \(ax + by = c\).

Worked example

Find both intercepts of the line \(2x + 3y = 12\), then describe how to sketch it.

1
For the \(y\)-intercept, set \(x = 0\): \(3y = 12\), so \(y = 4\). That gives \((0, 4)\).
2
For the \(x\)-intercept, set \(y = 0\): \(2x = 12\), so \(x = 6\). That gives \((6, 0)\).
3
Plot \((0, 4)\) and \((6, 0)\), then join them with a straight line.
\(y\)-intercept \((0, 4)\), \(x\)-intercept \((6, 0)\)

Check yourself

1. Where does \(y = 2x - 6\) cross the \(y\)-axis? +

Set \(x = 0\): \(y = 2(0) - 6 = -6\). The line crosses at \(\mathbf{(0, -6)}\).

2. Where does \(y = 2x - 6\) cross the \(x\)-axis? +

Set \(y = 0\): \(0 = 2x - 6\), so \(2x = 6\) and \(x = 3\). The line crosses at \(\mathbf{(3, 0)}\).

3. Find the \(y\)-intercept of \(x + 4y = 8\). +

Set \(x = 0\): \(4y = 8\), so \(y = 2\). The \(y\)-intercept is \(\mathbf{(0, 2)}\).


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