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Slope Intercept Form

One tidy equation, \(y = mx + c\), describes any non-vertical straight line. Learn to read the gradient \(m\) and the intercept \(c\) straight off it, and to build the equation back from a gradient and a point.

MYP 4Standard MathsCoordinate GeometryCriteria A · C · D~11 min read

Written by Fayek · MYP student

The form \(y = mx + c\)

Every straight line that is not vertical can be written as:

\[ y = mx + c \]

Here \(m\) is the gradient, which measures how steep the line is, and \(c\) is the \(y\)-intercept, the value of \(y\) where the line crosses the \(y\)-axis.

What each letter does
\(m\) is the change in \(y\) for every \(+1\) step in \(x\). A positive \(m\) slopes up to the right, a negative \(m\) slopes down. \(c\) shifts the whole line up or down without changing its steepness.

Reading \(m\) and \(c\)

When the equation is already in the form \(y = mx + c\), the numbers are sitting in plain view. In \(y = 3x - 1\), the gradient is \(3\) and the line crosses the \(y\)-axis at \((0, -1)\). The sign travels with the number: in \(y = -2x + 5\), the gradient is \(-2\) and the intercept is \(+5\).

The intercept keeps its sign

Read \(y = 4x - 7\) as gradient \(4\) and \(c = -7\), not \(+7\). The subtraction is part of the constant term.

Building the equation from a gradient and a point

If you know the gradient and one point on the line, substitute the point to find \(c\).

Worked example

A line has gradient \(4\) and passes through \((2, 5)\). Find its equation.

1
Start from \(y = mx + c\) with \(m = 4\): \(y = 4x + c\).
2
Substitute the point \((2, 5)\): \(5 = 4(2) + c\).
3
Solve: \(5 = 8 + c\), so \(c = -3\).
\(y = 4x - 3\)

Rearranging into the form

Sometimes an equation is given as \(ax + by = c\). Rearrange it so \(y\) is alone on the left, then read off \(m\) and \(c\) as before.

Worked example

Write \(4x + 2y = 10\) in slope intercept form and state its gradient and intercept.

1
Subtract \(4x\) from both sides: \(2y = -4x + 10\).
2
Divide every term by \(2\): \(y = -2x + 5\).
\(y = -2x + 5\), so gradient \(-2\) and \(y\)-intercept \((0, 5)\)

Check yourself

1. State the gradient and \(y\)-intercept of \(y = -3x + 7\). +

Read them off directly: the gradient is \(\mathbf{-3}\) and the \(y\)-intercept is \(\mathbf{(0, 7)}\).

2. A line has gradient \(5\) and passes through \((0, -2)\). Find its equation. +

The point \((0, -2)\) is the \(y\)-intercept, so \(c = -2\). With \(m = 5\), the equation is \(\mathbf{y = 5x - 2}\).

3. Rearrange \(3x + y = 4\) into slope intercept form. +

Subtract \(3x\) from both sides: \(y = -3x + 4\). So \(\mathbf{y = -3x + 4}\), gradient \(-3\), intercept \((0, 4)\).

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