Lesson Objectives
• Demonstrate an understanding of Linear Equations in Two Variables
• Demonstrate an understanding of slope
• Demonstrate an understanding of x and y-intercepts
• Learn how to write an equation in slope-intercept form
• Learn how to graph an equation from slope-intercept form
• Learn how to write an equation in point-slope form
• Learn how to write an equation in standard form

## Forms of a Line: Slope-Intercept Form, Point-Slope Form, & Standard Form

Over the course of the last few lessons, we have focused solely on linear equations in two variables. So far, we have learned how to generate and plot ordered pair solutions, graph linear equations, and find the slope for a linear equation in two variables. In this lesson, we will show how to algebraically manipulate a linear equation in two variables into three different forms: slope-intercept form, point-slope form, and standard form. Each form has a different use, but generally the most useful is slope-intercept form.

### Slope-Intercept Form

The Slope-Intercept form of a line is very useful. When a line is in slope-intercept form, we can obtain the slope and y-intercept by simple inspection. We can also quickly graph the equation. Slope-Intercept Form is obtained when we solve our linear equation in two variables for y.
Slope-Intercept Form:
y = mx + b
From our lesson on slope, we know that m stands for slope. This means if an equation is solved for y or in slope-intercept form, we can obtain the slope as the coefficient of the variable x. This may be significantly quicker versus using the slope formula. To prove this, suppose we have the following equation:
3x - 5y = 15
Two points on this line would be:
(0,-3) and (5,0)
Let's label (0,-3) as point 1 and (5,0) as point 2:
x1 = 0
y1 = -3
x2 = 5
y2 = 0
Now we plug into our slope formula: $$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ $$m = \frac{0 - (-3)}{5 - 0} = \frac{3}{5}$$ Our slope, m, is 3/5.
Now let's try to find our slope by putting our equation in slope-intercept form:
3x - 5y = 15
Solve for y:
-5y = -3x + 15
-5/-5 y = -3/-5 x + 15/-5
y = 3/5 x - 3
Since the coefficient of x is 3/5, this is our slope. We can see how much faster it is to find the slope from slope-intercept form vs using the slope formula.
Recall that we can find the y-intercept by replacing the x variable with a 0 and solving for y. When our equation is in slope-intercept form, notice how replacing the x with a 0 will always leave y = b:
y = mx + b
y = m(0) + b
y = b
This means our y-intercept can be found as (0,b). In our equation 3x - 5y = 15, which we solve for y as:
y = 3/5 x - 3
For this equation, we can see our y-intercept occurs at (0,3)
Let's look at a few examples.
Example 1: Find the slope and y-intercept for each
x - 5y = -10
Solve the equation for y:
-5y = -x - 10
-5/-5 y = -1/-5 x + -10/-5
y = 1/5 x + 2
slope = m = 1/5
y-intercept: (0,2)
Example 2: Find the slope and y-intercept for each
9x + 3y = 1
Solve the equation for y:
3y = -9x + 1
3/3 y = -9/3 x + 1/3
y = -3x + 1/3
slope = m = -3
y-intercept: (0, 1/3)
Example 3: Write the slope-intercept form of the line described
slope = 7, y-intercept: (0,-2)
y = mx + b
Plug in a 7 for m, this is our slope:
y = 7x + b
Plug in a (-2) for b, this is our y-intercept:
y = 7x - 2
A very useful application of slope-intercept form is graphing. In the past, we had to generate ordered pairs in order to sketch the graph of a linear equation in two variables. Now, we can use the y-intercept as a point, and the slope to plot additional points. Let's suppose we wanted to graph:
4x - 5y = -5
If we solve the equation for y and place it in slope-intercept form:
y = 4/5 x + 1
This means our slope is 4/5 and our y-intercept occurs at (0,1). Let's begin by plotting 0,1 on the coordinate plane: $$slope = \frac{4}{5} = \frac{rise}{run}$$ Since slope is rise over run, this tells us we will rise 4 units and run or move right 5 units to get to our next point (5,5): We could draw a line at this point since two points are all we need. Let's take this opportunity to show a little trick. Suppose we wrote our slope as: $$slope = \frac{-4}{-5} = \frac{rise}{run}$$ Since -4/-5 is the same as 4/5, we can use this form to find points in the other direction. Starting at our y-intercept, we can move 4 units down and 5 units left to arrive at (-5,-3): Now that we have plotted our points, let's graph our line: Let's take a look at an example.
Example 4: Graph each equation using slope-intercept form
4x - 3y = -9
Let's solve our equation for y:
-3y = -4x - 9
-3/-3 y = -4/-3 x + -9/-3
y = 4/3 x + 3
Our slope m is 4/3, our y-intercept occurs at (0,3). Let's plot the point (0,3) and gain two additional points using our slope 4/3: We can now sketch a line through our points: ### Point-Slope Form

In some cases, we will not be given enough information to write an equation in slope-intercept form. When we know one point on the line and the slope, we can use point-slope form to write the equation of the line.
Point-Slope Form: $$y - y_{1} = m(x - x_{1})$$ This form comes directly from our slope formula: $$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ Since we know one point on the line, we replace (x2,y2) with just x and y: $$m = \frac{y - y_{1}}{x - x_{1}}$$ (x,y) represents our known point, while (x1,y1) represents our unknown point. We can multiply both sides of the equation by:
(x - x1) to obtain our point-slope form: $$m \cdot (x - x_{1})= \frac{y - y_{1}}{x - x_{1}} \cdot (x - x_{1})$$ $$\require{cancel}m \cdot (x - x_{1})= \frac{y - y_{1}}{\cancel{x - x_{1}}} \cdot \cancel{(x - x_{1})}$$ $$y - y_{1} = m(x - x_{1})$$ Let's look at a few examples.
Example 5: Write the equation of the line described in point-slope form
through (3,4)
m = -2
We set up our point-slope formula:
y - y1 = m(x - x1)
(3,4) is our known point. We can plug in for x and y:
y - 4 = m(x - 3)
We know that m, our slope is -2:
y - 4 = -2(x - 3)
The equation above is our point-slope form. We can solve the equation for y and obtain slope-intercept form:
y = -2x + 6 + 4
y = -2x + 10
We know from this form that our line has a slope of (-2) and a y-intercept that occurs at: (0,10).
Example 6: Write the equation of the line described in point-slope form
through (-1,5) and (4,10)
Since we don't have the slope here, we first use the slope formula:
let (-1,5) be point 1, and (4,10) be point 2:
x1 = -1
y1 = 5
x2 = 4
y2 = 10
Slope Formula:
$$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$$ $$m = \frac{10 - 5}{4 - (-1)} = \frac{5}{5} = 1$$ Now that we know the slope, we can pick either given point and use point-slope form. Let's choose (-1,5) as our known point. Again, either point works here and will produce the same slope-intercept form.
We set up our point-slope formula:
y - y1 = m(x - x1)
Plug in our known point of (-1,5) and our known slope of 1:
y - 5 = 1(x - (-1))
We can solve for y and obtain slope-intercept form:
y - 5 = x + 1
y = x + 6
We know from this form that our line has a slope of 1 and a y-intercept that occurs at: (0,6)

### Standard Form

The standard form of a line is one that has a different definition between textbooks. Generally, we will see some format of:
ax + by = c
where a, b, and c are real numbers
In high school, we normally see a much stricter definition. In a lot of cases, we see that a, b, and c are required to be integers and a, the coefficient of x must be greater than 0. Let's follow this definition for now, and when we get into higher courses, we can relax our standard form definition. Let's take a look at an example.
Example 7: Write each line in standard form $$y = \frac{2}{5}x + 3$$ We want x and y on the left side: $$-\frac{2}{5}x + y = 3$$ We want the coefficient of x to be positive and an integer. We can multiply both sides of the equation by -5, to clear the denominator and make the coefficient of x positive. $$-5 \cdot -\frac{2}{5}x + -5 \cdot y = -5 \cdot 3$$ $$\require{cancel}\cancel{-5} \cdot \frac{2}{\cancel{-5}}x - 5y = -15$$ $$2x - 5y = -15$$