Lesson Objectives
• Demonstrate an understanding of slope-intercept form: y = mx + b
• Learn how to write the equation of a line given its slope and y-intercept
• Learn how to write the equation of a line in point-slope form
• Learn how to write the equation of a line using one point and the slope
• Learn how to write the equation of a line using only two points
• Learn how to write the equation of a line in standard form

## Slope-Intercept, Point-Slope, and Standard Forms of a Line

As we are studying linear equations in two variables, we need to learn how to algebraically manipulate our equation into different forms. In this lesson, we will learn how to change between slope-intercept form, point-slope form, and standard form. We will also learn how to obtain the equation of a line given its slope and a point on the line or simply two points on the line.

### Slope-Intercept Form

Over the course of the last few lessons, we have learned about the slope-intercept form of a line:
y = mx + b
When our linear equation in two variables is solved for y, it is in slope-intercept form. When an equation is in this format, we immediately know the slope (m) and y-intercept (b) from simple inspection. If we are given the task of writing the equation of a line, while being given the slope and y-intercept, we can plug in for m and b. Let's look at an example.
Example 1: Find the equation of the line with the given slope (m) and y-intercept (b).
slope » m » -4/3
y-intercept » (0,-5)
Plug in a (-4/3) for m and a (-5) for b in our slope-intercept form:
y = mx + b $$y=-\frac{4}{3}x - 5$$ Example 2: Find the equation of the line with the given slope (m) and y-intercept (b).
slope » m » -6
y-intercept » (0,13)
Plug in a (-6) for m and a (13) for b in our slope-intercept form:
y = mx + b $$y=-6x + 13$$

### Point-Slope Form

In some cases, we don't have enough information to write the equation of a line in slope-intercept form. When we know the slope, m, and one point on the line, we can use point-slope form.
Point-Slope Form:
y - y1 = m(x - x1)
When we look at the point-slope form of a line, we can see m, which again is used for slope. Additionally, we have x, y, x1, and y1. The x1 and y1 are used for our known point. As an example, suppose we had a line with a slope of 4 and a known point of (3,7). We can use this information to plug into the point-slope formula:
slope » m » 4
(x1,y1) » (3,7)
y - 7 = 4(x - 3)
Once we have the point-slope form of the line, we can solve for y and place the line in slope-intercept form:
y = 4x - 5
The point-slope form comes from a manipulation of our slope formula: $$m=\frac{y_{2}- y_{1}}{x_{2}- x_{1}}$$ Instead of the point (x2, y2), we will replace this notation with (x,y). Therefore (x,y) will be our unknown point and (x1, y1) will be our known point. $$m=\frac{y - y_{1}}{x - x_{1}}$$ If we multiply both sides by (x - x1): $$\require{cancel}m(x - x_{1})=\frac{y - y_{1}}{\cancel{(x - x_{1})}}\cdot \cancel{(x - x_{1})}$$ We obtain our point-slope form: $$y - y_{1}=m(x-x_{1})$$ Let's look at an example.
Example 3: Find the equation of the line described.
slope » m » 1/3
passes through the point » (-2,9)
Let's plug into our point-slope formula:
y - y1 = m(x - x1) $$y - 9=\frac{1}{3}\left(x - (-2)\right)$$ We can solve this equation for y and obtain slope-intercept form: $$y=\frac{1}{3}x + \frac{29}{3}$$ In some cases, we will only be given two points from the line. When this happens, we can use our slope formula to find the slope of the line and then use either point as (x1, y1) for use in the point-slope formula. Let's look at an example.
Example 4: Find the equation of the line described.
passes through the points: (1, 4), (7, -2)
Since we have only two points, we can't immediately use the point-slope formula. We first need to find the slope of the line using the slope formula. Let's let (1,4) be our first point and (7,-2) be our second point. $$m=\frac{-2 - 4}{7 - 1}=\frac{-6}{6}=-1$$ Now that we know our slope, we can take either known point and plug into the point-slope formula. Let's use (1,4) as our known point.
We will now plug into our point-slope formula:
y - y1 = m(x - x1) $$y - 4=-1(x - 1)$$ We can solve this equation for y and obtain slope-intercept form: $$y=-x + 5$$

### Standard Form of a Line

When we discuss the standard form of a line, note that the definition will vary based on the textbook and level of math. $$ax + by=c$$ Where a, b, and c are real numbers and a and b are not both zero.
Usually, in high school, we will see a stricter definition:
ax + by = c
• a, b, and c are integers
• a ≥ 0
• a and b are not both zero
• the GCF of a, b, and c is 1
Let's look at an example.
Example 5: Write each equation in standard form. $$y=\frac{3}{4}x + 6$$ First, let's move our term with the x variable to the left side: $$-\frac{3}{4}x + y=6$$ If we follow the stricter definition, we want our coefficient of x to be positive and an integer. We can multiply both sides of the equation by (-4): $$3x - 4y=-24$$

#### Skills Check:

Example #1

Write in slope-intercept form. $$(0, 3), m=5$$

A
$$y=5x - \frac{1}{3}$$
B
$$y=5x + 3$$
C
$$y=-5x - 3$$
D
$$y=-5x + 3$$
E
$$y=3x + 5$$

Example #2

Write in slope-intercept form. $$(1, 7), (5, 2)$$

A
$$y=-\frac{5}{4}x - 33$$
B
$$y=-\frac{4}{5}x + \frac{4}{33}$$
C
$$y=-4x + 5$$
D
$$y=-\frac{5}{4}x + \frac{33}{4}$$
E
$$y=5x + 6$$

Example #3

Write in standard form. $$(2, -4), (5, 7)$$

A
$$x + \frac{11}{3}y=\frac{34}{3}$$
B
$$2x - 5y=11$$
C
$$-x + 3y=4$$
D
$$11x - 3y=34$$
E
$$5x + 6y=7$$