# What is Slope Intercept Form?

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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)

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 yA = 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.