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# Writing a Linear Equation Using Slope and Points

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Author: Colleen Atakpu
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This lesson demonstrates how to write a linear equation using slope and points.

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Today we're gonna talk about writing linear equations given the slope of a line and or a point on that line. So we'll start by reviewing two of our forms of linear equations, slope-point form and slope-intercept form. And then we'll do some examples.

So let's start by reviewing two forms of linear equations. In slope-intercept form, the value of m is our slope and the value of b is our y-intercept. So to be able to write an equation in this form we need to know those two things about our line. And in slope-point form, the value of m is our slope and the values for x1 and y1 tell us a point that that line goes through. So to be able to write an equation in slope-point form, we need to know those two things about the line. So if we want to write an equation in either or both of these two forms, we need to be able to draw conclusions about the slope of the line, the intercepts of a line, and/or the points that it goes through.

So for this first example, let's see how we can write the equation of a line both in slope-point form and in slope-intercept form. So I know that the slope of my line is 2/3 and I also know that it passes through the point negative 9, 4. So because I know the slope of a line in a point that it passes through, I can write it in slope-point form.

So that is going to be in the form y minus y1 is equal to m times x minus x1, and I'm going to substitute my values into this equation. So my value for m is 2/3. My value for x1 is going to be negative 9. And my value for y1 is going to be 4. Now here in my parentheses I have x minus a negative 9, I'm going to go ahead and just simplify that right away. Subtracting a negative number is the same as adding a positive number, so I'm going to change this to x plus 9.

So now that I have my equation written in slope-point form, I can also go ahead and write it in slope-intercept form. To do that, I know that I want to make it look like y equals mx plus b which means I want to isolate this y variable on this side of the equation.

So I'm going to start by adding 4 to both sides of my equation. Here this will cancel, and I'll just be left with y is equal to 2/3 times x plus 9 plus 4. Now I'm going to simplify by distributing my 2/3 to both terms on the inside. So this is going to give me y is equal to 2/3 times x plus 2/3 times 9 is going to give me 6. And I have my plus 4 at the end. And then finally I can combine my two constant terms 6 and 4, which will give me 10. So I find that the equation of my line written in slope-intercept form is y is equal to 2/3 x plus 10. So I can see that the slope of my line is 2/3 and the y-intercept is 10.

So for my second example, I again I'm gonna write an equation both in slope-point form and in slope-intercept form. And for this example, I'm starting with two points that a line passes through. So I know that on a graph, this line passes through the points negative 4, 0 and negative 5, 3. Because I don't know the slope of the line and I need that both to write my equation in slope-point form and slope-intercept form, I'm gonna start by using these two points and the formula for slope to determine the slope.

So if you remember that the formula for slope is the difference in our y values, y2 and y1, over the difference in our x values, x2 and x1. So substituting my values from my points into my equation, I see that the difference in my y values is gonna be 3 minus 0. And the difference in my x values is gonna be negative 5 minus negative 4. Simplifying this, my numerator is going to be 3. And the denominator, negative 5 minus negative 4 is going to be negative 1. So that gives me a slope of 3 divided by negative 1 or negative 3.

So now that I know that my slope is negative 3, I can use either one of these two points. It doesn't matter which one I use to write my equation in slope-point form. So my equation in slope-point form is going to be y minus-- I'm gonna go ahead and use this y value of 0-- is equal to the slope of my line. So negative 3 times x minus my x value for that same point, so negative 4. So just like in our last example, I am subtracting a negative number. So to make this a little bit more simple, I'm just going to go ahead and change that to x plus 4.

So now I have this equation that is written in slope-point form, and I'm gonna convert it into slope-intercept form. So simplifying this. Because on this side of the equation I just have y minus 0, which is already just the same as y, I don't need to cancel out this number to get my y variable by itself. So y minus 0 is a ready just equal to y. And I've got negative 3 times x plus 4. So now all I need to do is distribute my negative 3 to both terms on the inside. Negative 3 times x will give me negative 3x and negative 3 times positive 4 will give me negative 12. So my equation written in slope-intercept form is negative y equals negative 3x minus 12. So I can see that my slope again is negative 3 and my y-intercept is negative 12.

So for my last example, I again am gonna write equations both in slope-point form and in slope-intercept form. But this time the information that I'm given is that the line that I want to write the equation for is perpendicular to the line with this equation y equals 3/5 x plus 4. And I also know that the line I want to write the equation for passes through the 3, 8.

So I'm gonna to start by figuring out the slope of my line. So because my line is perpendicular to the line with the equation y equals 3/5 x plus 4. I know that the slopes are gonna be opposite reciprocals of each other. So if the slope of this line is positive 3 over 5, I know that the slope of my line is gonna be negative-- because it's opposite-- and it's going to be 5 over three instead of 3 over 5, because 5 over 3 and 3 over 5 are reciprocals of each other.

So now that I know the slope of my line, I can then use the point that that line goes through to write the equation in slope-point form. So that's gonna look like y minus the y-coordinate, 8, is equal to my slope times x minus my x-coordinate, 3. Simplifying this, I'm gonna start by adding 8 to both sides so that I can isolate my y variable and I can write this equation in slope-intercept form. So adding 8 to both sides. Here this will cancel, and I'll be left with y is equal to negative 5 over 3 times x minus 3 plus 8.

And now I'm going to distribute my negative 5/3 to both terms on the inside. So this is gonna become y equals negative 5/3 x, and then negative 5/3 times a negative 3 is going to give me a positive 5. And then I'll bring down my plus 8. So lastly I just need to combine these two terms, and that will give me the equation y equals negative 5/3 x plus 13. So the slope of my line again is negative 5/3 and the y-intercept is 13.

So let's go over our key points from today. You can write an equation in slope-point form given certain information needed to determine the slope and a point on the line. You can write an equation in slope-intercept form given certain information needed to determine the slope and y-intercept of the line. Lines that are parallel to each other have the same slope. And finally, lines that are perpendicular to each other have slopes that are opposite reciprocals.

So I hope that these key points and examples helped you understand a little bit more about writing linear equations. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.

Formulas to Know
Slope

Slope-Intercept Form of a Line

Slope-Point Form of a Line