Standard Form Of A Line

You can represent any line that you can graph on a two-dimensional x-y axis by a linear equation. One of the simplest algebraic expressions, a linear equation is one that relates the first power of x to the first power of y. A linear equation can assume one of three forms: the slop-point form, the slope-intercept form and the standard form. You can write the standard form in one of two equivalent ways. The first is:

\(Ax + By + C = 0\)

where A, B and C are constants. The second way is:

\(Ax + By = C\)

Note that these are generalized expressions, and the constants in the second expression are not necessarily the same as those in the first one. If you want to convert the first expression to the second for particular values of A, B and C, you would have to write

\(Ax + By = -C\)

A linear equation defines a line on the x-y axis. Choosing any two points on the line, (x₁, y₁) and (x₂, y₂), allows you to calculate the slope of the line (m). By definition, it is the "rise over the run," or the change in the y-coordinate divided by the change in the x-coordinate.

\(m = \frac{∆y}{∆x} = \frac{y_2 – y_1}{ x_2 – x_1}\)

Now let (​x​₁, ​y​₁) be a particular point (​a​, ​b​) and let (​x​₂, ​y​₂) be undefined, that is be all values of ​x​ and ​y​. The expression for slope becomes

\(m = \frac{y – b}{x – a}\)

which simplifies to

\(m (x – a) = y – b\)

This is the slope point form of the line. If instead of (​a​, ​b​) you choose the point (0, ​b​), this equation becomes ​mx​ = ​y​ − ​b​. Rearranging to put ​y​ by itself on the left side gives you the slope intercept form of the line:

\(y = mx + b\)

The slope is usually a fractional number, so let it be equal to −​A​/​B​. You can then convert this expression to the standard form for a line by moving the ​x​ term and constant to the left side and simplifying:

\(Ax + By = C\)

where ​C​ = ​Bb​ or

\(Ax + By + C = 0\)

where ​C​ = −​Bb​

​Convert to standard form:​

\(y = \frac{3}{4}x + 2\)

1. Multiply Both Sides by 4

\(4y = 3x + 2\)

2. Subtract 3x from Both Sides

\(4y – 3x = 2\)

3. Multiply by

−1 to Make the x-Term Positive

\(3x – 4y = 2\)

This equation is in standard form. ​A​ = 3, ​B​ = −2 and ​C​ = 2

​Find the standard form equation of the line that passes through the points (-3, -2) and (1, 4).​

1. Find the Slope

\(\begin{aligned}
m &= \frac{y_2 – y_1}{x_2 – x_1}\)
\(&=\frac{1 – (-3)}{4 – 2} \
&= \frac{4}{ 2}\)
\(&= 2
\end{aligned}\)

2. Find Slope-Point Form Using Slope and One of the Points

The generic slope-point form is

\(m (x – a) = y – b\)

If you use the point (1, 4), this becomes

\(2 (x – 1) = y – 4\)

3. Simplify

\(2x – 2 – y + 4 = 0\)
\(2x – y + 2 = 0\)

This equation is in standard form ​Ax​ + ​By​ + ​C​ = 0 where ​A​ = 2, ​B​ = −1 and ​C​ = 2