Year 11 General Linear Equations And Their Graphs

The Slope–Intercept Form Of The Equation Of A Straight Line

11 practice questions 1 video lesson Theory + worked examples

Theory

The slope-intercept form of a straight line is \(y = mx + c\), where \(m\) is the slope and \(c\) is the \(y\)-intercept. This page covers reading off \(m\) and \(c\), rearranging equations into this form, and building the equation from a slope and intercept or from two points.

The slope-intercept form of a straight-line equation is \(y = mx + c\), where \(m\) is the slope (gradient) and \(c\) is the \(y\)-intercept — the value of \(y\) when \(x = 0\), i.e.\ where the line crosses the \(y\)-axis.

When an equation is already in this form, you can read off \(m\) and \(c\) immediately: the slope is the coefficient of \(x\) (with its sign), and the \(y\)-intercept is the constant term. For example, \(y = 3x - 4\) has slope \(3\) and \(y\)-intercept \(-4\).

If the equation is given in another form (such as \(ax + by + c = 0\)), solve for \(y\) to convert. To build the equation from a slope and intercept, just substitute into \(y = mx + c\). From a graph or two points, read off \(c\) and compute \(m\) from rise over run.

A line \(y = mx + c\) crosses the \(y\)-axis at \((0, c)\); the slope \(m\) is rise over run.

In a real-world phone plan, \(c\) is the fixed fee and \(m\) is the cost per extra minute.

The slope-intercept form:

\[y = mx + c\]

y = m x + c

where:

Symbol	Meaning
\(m\)	the slope (gradient); rate of change of \(y\) per unit of \(x\)
\(c\)	the \(y\)-intercept; the value of \(y\) when \(x = 0\)

Reading off \(m\) and \(c\) from common equations:

Equation	Slope \(m\)	\(y\)-intercept \(c\)
\(y = 3x - 4\)	\(3\)	\(-4\)
\(y = -\dfrac{2}{3}x + 6\)	\(-\dfrac{2}{3}\)	\(6\)
\(y = 5\) (no \(x\) term)	\(0\)	\(5\)

Real-world meaning. For a phone plan \(C = 0.15t + 25\): the slope \(0.15\) is the cost per extra minute (rate of change) and the intercept \(25\) is the fixed monthly fee (cost when \(t = 0\)).

How to build a line equation from a graph or two points

Read the \(y\)-intercept \(c\) directly from where the line crosses the \(y\)-axis.
Compute the slope using \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\) for any two points on the line.
Substitute \(m\) and \(c\) into \(y = mx + c\).

Rearranging. If the equation is in general form \(ax + by + c = 0\), or in point-slope form \(y - y_1 = m(x - x_1)\), solve for \(y\) to get into slope-intercept form. From there both \(m\) and \(c\) are easy to read off.

Example 1 — Read Off m and c

For the line \(y = -\dfrac{3}{4}x + 2\), state the slope and \(y\)-intercept.

Solution

The slope is the coefficient of \(x\); the \(y\)-intercept is the constant.

slope	\(=\)	\(-\dfrac{3}{4}\)
\(y\)-intercept	\(=\)	\(2\)

m = - \frac{3}{4}, c = 2

Example 2 — Build from m and c

A line has slope \(-3\) and \(y\)-intercept \(7\). Write its equation.

Solution

Substitute \(m = -3\) and \(c = 7\) into \(y = mx + c\).

\(y\)	\(=\)	\(mx + c\)
\(y\)	\(=\)	\(-3x + 7\)

y = - 3 x + 7

Example 3 — Rearrange to find m, c

Find the slope and \(y\)-intercept of \(2x + 3y = 12\).

Solution

Solve for \(y\) by moving the \(2x\) across and dividing by \(3\).

\(2x + 3y\)	\(=\)	\(12\)
\(3y\)	\(=\)	\(-2x + 12\)
\(y\)	\(=\)	\(-\dfrac{2}{3}x + 4\)

Slope \(m = -\dfrac{2}{3}\); \(y\)-intercept \(c = 4\).

y = - \frac{2}{3} x + 4

Example 4 — Equation from Two Points

A line passes through \((0, 1)\) and \((3, 7)\). Find its equation in slope-intercept form.

Solution

\(y\)-intercept \(c = 1\) (read directly from \((0, 1)\)). Find the slope.

\(m\)	\(=\)	\(\dfrac{7 - 1}{3 - 0}\)
\(m\)	\(=\)	\(\dfrac{6}{3} = 2\)

So \(y = 2x + 1\).

y = 2 x + 1

Common pitfalls

Ignoring the sign of the slope. The coefficient of \(x\) is the slope, including its sign. \(y = -2x + 5\) has slope \(-2\), not \(2\).

Sign flip when rearranging. \(2x + 3y = 6\) becomes \(3y = -2x + 6\) — the \(2x\) becomes \(-2x\) when it crosses the equals sign. Then divide every term by \(3\): \(y = -\dfrac{2}{3}x + 2\).

Wrong intercept. The \(y\)-intercept is the value of \(y\) when \(x = 0\), so it's the constant at the end of \(y = mx + c\), not the coefficient of \(x\). Don't confuse \(m\) and \(c\).

Dividing only some terms. When isolating \(y\), divide every term on the right by the coefficient of \(y\). \(3y = -2x + 12\) gives \(y = -\dfrac{2}{3}x + 4\) — both \(-2x\) and \(12\) get divided.

Frequently asked questions

What is the slope-intercept form of a line?

The slope-intercept form is y equals m x plus c, where m is the slope (gradient) and c is the y-intercept — the value of y when x equals 0, which is where the line crosses the y-axis.

How do I read the slope and y-intercept from an equation in y equals mx plus c form?

The slope is the coefficient of x (including its sign) and the y-intercept is the constant term. For y equals 3x minus 4, the slope is 3 and the y-intercept is minus 4. For y equals minus 2 over 3 x plus 6, the slope is minus 2 over 3 and the y-intercept is 6.

How do I rearrange an equation like 2x plus 3y equals 6 into slope-intercept form?

Solve for y. Move the x term to the other side, then divide every term by the coefficient of y. So 2x plus 3y equals 6 becomes 3y equals minus 2x plus 6, then y equals minus 2 over 3 x plus 2.

How do I build the equation from a slope and a y-intercept?

Just substitute into y equals m x plus c. If the slope is 4 and the y-intercept is minus 3, the equation is y equals 4x minus 3.

How do I find the equation of a line from two points?

If one of the points is on the y-axis, the y-intercept c is the y value of that point. Find the slope using m equals (y2 minus y1) over (x2 minus x1). Then substitute m and c into y equals m x plus c.

What does m and c mean in a real-world problem?

The slope m is the rate of change (cost per minute, dollars per kilometre, etc). The y-intercept c is the fixed starting amount (monthly fee, flagfall, initial deposit) — the value when x equals 0.