Standard form of a line

Why it survives alongside the friendlier form

Slope-intercept form is easier to graph, so standard form needs a reason to exist. It has two.

It can write a vertical line. Set $B = 0$ and you get $Ax = C$, which is $x = k$. No value of $m$ can do that, because a vertical line has undefined slope.

It lines up for elimination. In a system of equations, matching coefficients sit in columns and add away cleanly.

Both intercepts in two steps

Set each variable to zero in turn.

$$y = 0 \;\Rightarrow\; x = \frac{C}{A} \qquad x = 0 \;\Rightarrow\; y = \frac{C}{B}$$

For $3x + 4y = 12$: the $x$-intercept is $(4, 0)$ and the $y$-intercept is $(0, 3)$. Plot both and draw the line — often faster than converting.

Reading the slope without converting

Solve the general form for $y$:

$$By = -Ax + C \quad\Rightarrow\quad y = -\frac{A}{B}x + \frac{C}{B}$$

So the slope is $-\tfrac{A}{B}$ and the $y$-intercept is $\tfrac{C}{B}$, whenever $B \ne 0$.

For $3x + 4y = 12$ that gives a slope of $-\tfrac{3}{4}$ — no rearranging needed.

Worked examples

Convert $y = \tfrac{2}{3}x - 4$ to standard form.

Clear the fraction by multiplying through by $3$, then move the $x$ term across:

$$3y = 2x - 12 \quad\Rightarrow\quad -2x + 3y = -12 \quad\Rightarrow\quad 2x - 3y = 12$$

The last step multiplies by $-1$ so that $A$ is positive, which is the usual convention.

Convert $2x - 5y = 20$ to slope-intercept form.

$$-5y = -2x + 20 \quad\Rightarrow\quad y = \tfrac{2}{5}x - 4$$