Finding Slope-Intercept Equation
Hello! Today we are going to take a look at how to find the slope-intercept form of a linear equation when given a point on the line and the slope of the line, and when we are given two points on a line. Remember, the slope-intercept form of a linear equation is:
\(y=mx+b\)
where \(m\) stands for our slope, and \(b\) stands for the \(y\)-intercept.
Let’s start with finding the equation when given a point on the line and the slope of the line. So, let’s say we’re given the point \((3,5)\).
Point: \((3,5)\)
This means that the point \((3,5)\) is a point that’s on the line. And let’s say that we’re also given that our slope of the line is 2.
Slope: \(2\)
So, the first thing we’re going to do is write out the slope-intercept form equation of the line:
\(y=mx+b\)
This helps us remember the formula, and it makes sure we plug in values into the correct places. Now, to create this equation, we need to plug in values for both \(m\) and \(b\). Remember, \(m\) stands for the slope, which we are given. The slope is 2. So we’re going to plug in this value.
\(y=2x+b\)
Now we need to find \(b\). To find \(b\), we are going to plug in the point we are given for \(x\) and \(y\). The point \((3,5)\) stands for \((x,y)\), so \(x=3\) and \(y=5\). So, we’re going to plug in these values and then solve for \(b\).
\(5=2(3)+b\)
And now we solve for \(b\). Start by multiplying 2 and 3 on the right side.
\(5=6+b\)
Next, subtract 6 from both sides.
\(-1=b\)
So, now that we know that our \(y\)-intercept is \(-1\), we can go back to our equation over here and plug in this value.
\(y=2x+(-1)\)
Now remember, adding a negative number is the same thing as subtracting, so we can rewrite this as:
\(y=2x-1\)
And that’s our final answer! Let’s try another one.
Find the slope-intercept equation of a line that passes through the point \((-6,2)\) and has a slope of 5.
First, write out the general form of a slope-intercept equation.
\(y=mx+b\)
Then, plug in the slope. Our slope is 5, so we’ll get:
\(y=5x+b\)
Now we’re going to plug in our point for \(x\) and \(y\).
\(2=5(-6)+b\)
Now, solve for \(b\). Start by multiplying 5 and -6.
\(2=-30+b\)
And now we’ll add 30 to both sides of our equation. That gives us:
\(32=b\)
So our \(y\)-intercept is 32. Now we can go back here and plug this value in for \(b\).
\(y=5x+32\)
Now that we’ve got the hang of that, let’s take a look at how to find the slope-intercept equation when we’re given two points on the line. So, let’s say we’re given the points:
Points: \((6,5)\) and \((7,1)\)
First, write out the general form of a slope-intercept equation.
\(y=mx+b\)
Now we need to plug in values for \(m\) and \(b\). This time we aren’t given \(m\), but we can solve for it using the slope formula.
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Let’s use the point \((6,5)\) as \((x_{1},y_{1})\) and the point \((7,1)\) as \((x_{2}, y_{2})\). If you switched the points and said that \((6,5)\) was \((x_{2},y_{2})\) and \((7,1)\) was \((x_{1},y_{1})\), that would also be okay, you’ll get the same slope value either way. Now, let’s plug in our values and find our slope.
\(m=\frac{1-5}{7-6}=\frac{-4}{1}=-4\)
So our slope \((m)\) is \(-4\). Now that we know this, we can follow the same steps as we did in our previous practice problems. Plug in \(-4\) into the equation for \(m\).
\(y=-4x+b\)
For this next part, where we plug in a point, you can use either point. If you do it correctly, you will get the same answer both ways. For this example, let’s use the point \((6,5)\).
\(5=-4(6)+b\)
Now we’re going to start by multiplying \(-4\) and \(6\).
\(5=-24+b\)
Then, we’ll add 24 to both sides.
\(29=b\)
So, since our \(y\)-intercept, \(b\), is equal to 29, we can now go over here and plug that into our equation. So we’ll have:
\(y=-4x+29\)
And that’s our final answer! Before we go, I want to try one more problem.
Find the point-slope equation of a line that passes through the points \((-9,11)\) and \((12,18)\).
First things first, write out your general equation.
\(y=mx+b\)
Then, we solve for \(m\) using the slope formula.
\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
If we plug in our points, we’ll get:
\(m=\frac{18-11}{12-(-9)}\)
We can rewrite this denominator as, \(12+9\), since subtracting a negative number is the same as adding that number.
\(m=\frac{18-11}{12+9}=\frac{7}{21}\)
Then we can simplify this fraction by dividing both the numerator and denominator by 7 and get:
\(m=\frac{1}{3}\)
So, our slope is equal to \(\frac{1}{3}\). Now that we have our slope, we can plug it into our equation.
\(y=\frac{1}{3}x+b\)
Next, choose either point to plug in for \(x\) and \(y\). This time, let’s use the point \((12,18)\).
\(18=\frac{1}{3}(12)+b\)
Now we solve for \(b\). Start by multiplying 13 and 12.
\(18=4+b\)
And then we’ll subtract 4 from both sides.
\(14=b\)
So, now that we know that our \(y\)-intercept is equal to 14, we can come back over here and plug it into our equation and get:
\(y=\frac{1}{3}x+14\)
And that’s all there is to it! I hope this video helped you learn how to find the slope-intercept form of a linear equation. Thanks for watching, and happy studying!
Slope-Intercept Equation Practice Questions
Use the slope-intercept form to write the equation of a line that has a slope of –2 and contains the point \((4,-3)\).
\(y=-2x-5\)
\(y=-2x-7\)
\(y=-2x+5\)
\(y=-2x-11\)
The equation of a line in slope-intercept form is:
\(y=mx+b\)
- \(m\) is the slope of the line.
- \(b\) is the \(y\)-intercept of the line.
- \((x,y)\) is a point on the line.
First, substitute \(m=-2\) into the equation.
\(y=-2x+b\)
Next, substitute the coordinates of the given point, \(x=4\) and \(y=-3\) into the equation and solve for \(b\).
\(-3=-2\left(4\right)+b\)
\(-3=-8+b\)
\(-3+8=-8+b+8\)
\(5=b\)
Lastly, substitute the value of \(b\) into the equation to write the equation of the line in slope-intercept form.
\(y=-2x+5\)
Use the slope-intercept form to write the equation of a line that contains the points \((1,5)\) and \((3,9)\).
\(y=\frac{1}{2}x-\frac{11}{2}\)
\(y=\frac{1}{2}x+\frac{9}{2}\)
\(y=2x-7\)
\(y=2x+3\)
The equation of a line in slope-intercept form is:
\(y=mx+b\)
- \(m\) is the slope of the line.
- \(b\) is the \(y\)-intercept of the line.
- \((x,y)\) is a point on the line.
First, find the slope of the line using the given points. The equation for slope is:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute the known information and solve for \(m\) to get the following equation.
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{9-5}{3-1}=\frac{4}{2}=2\)
Now, substitute the slope into the slope-intercept form.
\(y=2x+b\)
Next, substitute the coordinates of either point into the equation, then solve for \(b\). Substitute the first point with \(x=1\) and \(y=5\), we get the following equation.
\(5=2\left(1\right)+b\)
\(5=2+b\)
\(5-2=2+b-2\)
\(3=b\)
Lastly, substitute the value of \(b\) into the equation to write the equation of the line in slope-intercept form.
\(y=2x+3\)
Use the slope-intercept form to write the equation of a line that contains the points \((-6,13)\) and \((12,16)\).
\(y=\frac{1}{6}x+14\)
\(y=\frac{1}{6}x-18\)
\(y=6x-56\)
\(y=6x+88\)
The equation of a line in slope-intercept form is:
\(y=mx+b\)
- \(m\) is the slope of the line.
- \(b\) is the \(y\)-intercept of the line.
- \((x,y)\) is a point on the line.
First, find the slope of the line using the given points. The equation for slope is:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute the known information and solve for \(m\).
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{16-13}{12-\left(-6\right)}=\frac{3}{12+6}=\frac{3}{18}=\frac{1}{6}\)
Now, substitute the slope into the slope-intercept form.
\(y=\frac{1}{6}x+b\)
Next, substitute the coordinates of either point into the equation, then solve for \(b\). Substitute the second point with \(x=12\) and \(y=16\), to get the following equation.
\(16=\frac{1}{6}\left(12\right)+b\)
\(16=2+b\)
\(16-2=2+b-2\)
\(14=b\)
Lastly, substitute the value of \(b\) into the equation to write the equation of the line in slope-intercept form.
\(y=\frac{1}{6}x+14\)
You run every day for a month to get in shape. The third day you ran 3 miles, and the sixth day you ran 5 miles. Suppose the number of miles you run each day can be modeled by a linear equation where \(x\) is the day of the month, and \(y\) is the number of miles you ran on a particular day of the month. Which of the following is the linear equation that models the number of miles you run on a particular day of the month?
\(y=\frac{2}{3}x-9\)
\(y=\frac{2}{3}x+1\)
\(y=\frac{3}{2}x-4\)
\(y=\frac{3}{2}x-14\)
Since \(x\) is the day of the month and \(y\) is the number of miles you ran on that day, we can write the ordered pairs \((3,3)\) and \((6,5)\) to represent the number of miles ran on the third and sixth days of the month, respectively.
The equation of a line in slope-intercept form is:
\(y=mx+b\)
- \(m\) is the slope of the line.
- \(b\) is the \(y\)-intercept of the line.
- \((x,y)\) is a point on the line.
First, find the slope of the line using the given points. The equation for slope is:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute the known information and solve for \(m\).
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{5-3}{6-3}=\frac{2}{3}\)
Now, substitute the slope into the slope-intercept form.
\(y=\frac{2}{3}x+b\)
Next, substitute the coordinates of either point into the equation, then solve for \(b\). Substitute the second point with \(x=6\) and \(y=5\), to get the following equation.
\(5=\frac{2}{3}\left(6\right)+b\)
\(5=4+b\)
\(5-4=4+b-4\)
\(1=b\)
Lastly, substitute the value of \(b\) into the equation to write the equation of the line in slope-intercept form.
\(y=\frac{2}{3}x+1\)
A tree is planted in a park. Each week for a year the height of the tree is measured to see how much it has grown. On the fifth week, the tree measures 10 feet tall, on the tenth week the tree measures 14 feet tall. Suppose the height of the tree can be modeled by a linear equation where \(x\) represents a week of the year, and \(y\) is the height of the tree at week \(x\). Which of the following is the linear equation that models the height of the tree in terms of the week of the year?
\(y=\frac{5}{4}x+\frac{15}{4}\)
\(y=\frac{5}{4}x-\frac{65}{4}\)
\(y=\frac{4}{5}x+6\)
\(y=\frac{4}{5}x-14\)
Since \(x\) is a week of the year, and \(y\) is the height of the tree on that week of the year, we can write the ordered pairs \((5,10)\) and \((10,14)\) to represent the height of the tree on the fifth and tenth weeks of the year, respectively.
The equation of a line in slope-intercept form is:
\(y=mx+b\)
- \(m\) is the slope of the line.
- \(b\) is the \(y\)-intercept of the line.
- \((x,y)\) is a point on the line.
First, find the slope of the line using the given points. The equation for slope is:
\(m=\frac{y_2-y_1}{x_2-x_1}\)
where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute the known information and solve for \(m\).
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{14-10}{10-5}=\frac{4}{5}\)
Now, substitute the slope into the slope-intercept form.
\(y=\frac{4}{5}x+b\)
Next, substitute the coordinates of either point into the equation, then solve for b. Substitute the first point with \(x=5\) and \(y=10\), to get the following equation.
\(10=\frac{4}{5}\left(5\right)+b\)
\(10=4+b\)
\(10-4=4+b-4\)
\(6=b\)
Lastly, substitute the value of \(b\) into the equation to write the linear equation of the line in slope-intercept form.
\(y=\frac{4}{5}x+6\)