The equation of a straight line

There are infinitely many straight lines which all share the same slope, so how do we identify a specific one? We fix one point on the line: only one line with that slope passes through it.

The most convenient point to fix is where the line crosses the $y$-axis. Every point on the $y$-axis has $x$-coordinate $0$, so we write this point as $(0, c)$, where $c$ is the height of the crossing.

In the diagram below, the slider sets the slope shared by all the faint lines, and the point on the $y$-axis picks out one of them.

Reading $m$ and $c$ off an equation tells us which line it describes. In $y = 3x - 2$ we have $m = 3$ and $c = -2$, so the line rises and crosses the $y$-axis at $(0, -2)$.

When there is no constant term, as in $y = 4x$, we have $c = 0$, so the line passes through the origin.

Four straight lines are drawn on the axes below. Which of them is $y = -\dfrac{1}{2}x + 3$?

Solution

In $y = -\dfrac{1}{2}x + 3$ we have $c = 3$, so the line we want crosses the $y$-axis at $(0, 3)$. Lines $A$, $B$ and $C$ all pass through that point, while $D$ crosses at $(0, -2)$, so $D$ is ruled out.

Next we use the slope. Here $m = -\dfrac{1}{2}$, which is negative, so the line falls from left to right. Line $B$ rises, so $B$ is ruled out too.

That leaves $A$ and $C$, and we separate them by the size of their slope. Starting from $(0, 3)$, line $A$ drops $1$ unit for every $2$ units across, so its slope is $-\dfrac{1}{2}$. Line $C$ drops $2$ units for every $1$ unit across, giving a much steeper $-2$.

The line is $A$.

Writing the equation of a drawn line means finding the two numbers in $y = mx + c$ from the picture. The diagram below marks where each one comes from: the point $(0, c)$ where the line crosses the $y$-axis gives $c$, and the rise and run from $(0, c)$ to a second point on the line give the slope $m$.

Write the equation of the straight line drawn below.

Solution

1. The line crosses the $y$-axis at $(0, -1)$, so $c = -1$.

2. We take the $y$-axis crossing as $(x_1, y_1) = (0, -1)$ and the point $(1, 1)$, where the line passes exactly through a grid point, as $(x_2, y_2)$. The green legs show the run and rise between them:

   $$\begin{align*} m &= \dfrac{y_2 - y_1}{x_2 - x_1} \\ &= \dfrac{1 - (-1)}{1 - 0} \\ &= \dfrac{2}{1} \\ &= 2 \end{align*}$$

3. Writing $m = 2$ and $c = -1$ into $y = mx + c$ gives the equation

   $$y = 2x - 1$$

A horizontal line stays at one height, so every point on it has the same $y$-coordinate. The horizontal line through $(2, 1)$ contains $(-3, 1)$, $(0, 1)$, $(4, 1)$ and every other point with $y$-coordinate $1$, whatever its $x$-coordinate. Its equation is $y = 1$.

A vertical line stays at one position across, so every point on it has the same $x$-coordinate. The vertical line through $(2, 1)$ is $x = 2$.

A horizontal line fits the slope-intercept form with $m = 0$, since $y = c$ is $y = 0x + c$. A vertical line does not: the equation $x = k$ contains no $y$, so it cannot be rearranged into $y = mx + c$.

Write the equations of the horizontal line and the vertical line through $(-4, 3)$. Both lines are drawn below.

Solution

Here the point is $(a, b) = (-4, 3)$.

• For the blue horizontal line we keep the point's $y$-coordinate, $3$, and leave $x$ free. Its equation is $y = 3$.
• For the red vertical line we keep the point's $x$-coordinate, $-4$, and leave $y$ free. Its equation is $x = -4$.

We check against the point: $(-4, 3)$ has $y$-coordinate $3$, so it satisfies $y = 3$, and it has $x$-coordinate $-4$, so it satisfies $x = -4$.