Linear equations come in three basic forms: point-slope, standard and slope-intercept. The general format of slope-intercept is *y* = *Ax* + *B*, where *A* and *B* are constants. Although the different forms are equivalent, providing the same results, the slope-intercept form quickly gives you valuable information about the line it produces.

#### TL;DR (Too Long; Didn't Read)

**TL;DR (Too Long; Didn't Read)**

The slope-intercept form of a line is *y* = *Ax* + *B*, where *A* and *B* are constants and *x* and *y* are variables.

## Slope-Intercept Breakdown

The slope-intercept form, *y* = *Ax* + *B* has two constants, *A* and *B*, and two variables, *y* and *x*. Mathematicians call *y* the dependent variable because its value depends on what happens on the other side of the equation. The *x* is the independent variable because the rest of the equation depends on it. The constant *A* determines the slope of the line and *B* is the value of the *y*-intercept.

## Slope and Intercept Defined

The slope of a line reflects the line’s “steepness,” and if it increases or decreases. To give some examples, a horizontal line has a slope of zero, a gently rising line has a slope with a small numeric value, and a steeply rising line has a slope with a large value. The fourth type of slope is undefined; it is vertical. The sign of the slope shows whether the line rises or falls in value going from left to right. A positive slope means the line rises, and a negative slope means it falls.

The intercept is the point at which the line crosses the *y*-axis. Going back to the form, *y* = *Ax* + *B*, you can find the point by taking the value of *B* and finding that number on the *y* axis, where *x* is zero. For example, if your line equation is *y* = 2_x_ + 5, the point lies at (0, 5), right on the *y* axis.

## Two Other Forms

In addition to the slope-intercept form, two other forms are in common use, standard and point-slope. The standard form of a line is *Ax* + *By* = *C*, where *A*, *B* and *C* are constants. For example, 10_x_ + 2_y_ = 1 describes a line in this form. The point-slope form is *y* − *A* = *B*(*x −* *C*). This equation provides an example of the point slope form: *y −* 2 = 5(*x −* 7).

## Graphing with Slope-Intercept

You need two points to draw a line on a graph. The slope-intercept form gives you one of those points automatically — the intercept. Plot the first point using the value of *B* following the directions described above. Finding the second point takes a little algebra work. In your line equation, set the value of *y* to zero, then solve for *x*. For example, using *y* = 2_x_ + 5, solve 0 = 2_x_ + 5 for *x*:

Subtracting 5 from both sides gives you −5 = 2_x_.

Dividing both sides by 2 gives you −5 ÷ 2 = *x*.

Mark the point at ( −5/2, 0). You already have a point at (0, 5). Using a ruler, draw a line connecting the two points.

## Finding Parallel Lines

Creating a line parallel to one written as slope-intercept is simple. Parallel lines have the same slope but different *y*-intercepts. So simply keep the slope variable *A* from your original line equation and use a different variable for *B*. For example, to find a line parallel to *y* = 3.5_x_ + 20, keep 3.5_x_ and use a different number for *B*, such as 14, so the equation for the parallel line is *y* = 3.5_x_ + 14. You may also need to find a line that passes through a particular point at (*x*, *y*). For this exercise, plug in the values of *x* and *y* and solve for the *y*-intercept, *B*. For example, you want to find the line that passes through the point (1, 1). Set *x* and *y* to the values of the point given and solve for *B*:

Substitute the point values for *x* and *y*:

1 = 3.5 × 1 + *B*

Multiply the *x* value (1) by the slope (3.5):

1 = 3.5 + *B*

Subtract 3.5 from both sides:

1 − 3.5 = *B*

−2.5 = *B*

Plug the value of *B* into your new equation.

*y* = 3.5_x −_ 2.5

## Finding Perpendicular Lines

Perpendicular lines cross one another at right angles. To do that, the slope of of the perpendicular line is −1 / *A* of the original line, or negative one divided by the original slope. To find a line perpendicular to *y* = 3.5_x_ + 20, divide −1 by 3.5 and get the result, −2/7. Any line with the slope of −2/7 will be perpendicular to *y* = 3.5_x_ + 20. To find a perpendicular line that passes through a given point (*x*, *y*), plug the values of *x* and *y* into your equation and solve for the *y*-intercept, *B*, as above.