Equations of Parallel Lines

Hello! Today we are going to take a look at how to find the equation of a line that passes through a given point and is parallel to another line. Let’s start with an example.

Find the equation of a line that passes through the point \((2,-1)\) and is parallel to the line \(y=3x+4\).

Remember, parallel lines have the same slope, so we need to start by finding the slope of the line we are given. The line we are given is in slope-intercept form, \(y=mx+b\), where \(m\) is the slope of the line. So the slope of the line \(y=3x+4\) is 3.

\(m=3\)

 

Now that we know the slope of the line and a point that the line passes through \((2,-1)\), we can plug these values into either our general slope-intercept equation or our general point-slope equation. For this example, we’ll use the slope-intercept equation.

\(y=mx+b\)

 

We know our value for \(m\) and we know values for \(x\) and \(y\) that make this equation true (because the line passes through them), so we can plug in these three values in and solve for \(b\).

\(-1=3(2)+b\)

 

First, multiply.

\(-1=6+b\)

 

Then, subtract 6 from both sides.

\(-7=b\)

 

Now that we know what \(b\) is, we can use our values for \(m\) and \(b\) to create the equation for our line. Our line in slope-intercept form is:

\(y=3x+(-7)\)

 

This simplifies to:

\(y=3x-7\)

 

And there we have it! The equation of a line that passes through the point \((2,-1)\) and is parallel to the line \(y=3x+4\) is \(y=3x-7\).

Let’s try another example.

Find the equation of a line that passes through the point \((4,6)\) and is parallel to the line \(y=-2x-1\).

First, find the slope of the new line. Since parallel slopes are the same, the slope of our new line is also \(-2\).

\(m=-2\)

 

For this example, we are going to find our equation by plugging in our slope and point into the point-slope equation.

\(y-y_{1}=m(x-x_{1})\)

 

Remember, since this form is created to tell you a specific point on the line, we will plug in our point \((4,6)\) for \((x_{1},y_{1})\) instead of \((x,y)\) like in the last example. Let’s plug in the values that we know.

\(y-6=-2(x-4)\)

 

Now, if we want our answer in point-slope form, we can stop here. If we want it in slope-intercept form, we just have to do a little bit of rearranging. First, distribute on the right side, and then move the \(-6\) to the right side.

\(y-6=-2x+8\)
 
\(y=-2x+14\)

 

So our answer is \(y=-2x+14\). This is the line that is parallel to the line, \(y=-2x-1\), and passes through the point \((4,6)\).

Let’s try one more example together before we go.

Find the equation of a line that passes through the point \((-3,8)\) and is parallel to the line \(y=7x-16\).

First, determine the slope of the new line. It will be the same as the slope of the given line, so the slope is 7.

\(m=7\)

 

Then, plug in your slope into either the slope-intercept equation or the point-slope equation. We’ll work through both ways for this example, starting with the slope-intercept equation. First, write out your slope-intercept equation.

\(y=mx+b\)

 

Then we’re going to plug in our values for \(m\), \(x\), and \(y\).

\(8=7(-3)+b\)

 

So to solve, we’re going to start by multiplying 7 and \(-3\).

\(8=-21+b\)

 

Then we add 21 to both sides.

\(29=b\)

 

Now that we know our value for \(b\), plug it into the slope-intercept equation, along with the value for \(m\). So our final equation is:

\(y=7x+29\)

 

If we wanted to use the point-slope equation instead, we would get the same answer.

\(y-y_{1}=m(x-x_{1})\)

 

Plug in the values for \(m\), \(x_{1}\), and \(y_{1}\).

\(y-8=7(x-(-3))\)

 

If we simplify the subtracting a negative part, we get:

\(y-8=7(x+3)\)

 

Now, we can distribute and rearrange to convert our equation to slope-intercept form.

\(y-8=7x+21\)
 
\(y=7x+29\)

 

Sure enough, that’s the exact same equation that we got before, so we know that our answer is correct either way that we choose to solve this problem.

I hope that this video has been helpful. Thanks for watching, and happy studying!

Practice Questions

Question #1:

 
Find the equation of the line that contains the point \((3,-5)\) and is parallel to the line \(y=2x+1\).

\(y=-2x+1\)
\(y=2x-11\)
\(y=x-8\)
\(y=-x+1\)
Answer:

Lines are parallel if they have the same slope. The given line is in the form \(y=mx+b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. For the given line, the slope is \(m=2\), so the slope of the parallel line is also \(m=2\).

We can substitute the given point and the slope into the slope-intercept form of a line to find the equation for the line containing the point \((3,-5)\). Substitute the values into slope-intercept form, then solve for \(b\).

\(y=mx+b\)
\(-5=2(3)+b\)
\(-5=6+b\)
\(-5-6=6+b-6\)
\(-11=b\)

Plug \(b=-11\) and \(m=2\) into the slope-intercept form of a line, the equation of the line containing the point \((3,-5)\) is:

\(y=mx+b\)
\(y=(2)x+(-11)\)
\(y=2x-11\)

Question #2:

 
Find the equation of the line that contains the point \((2,11)\) and is parallel to the line \(y=3x-10\).

\(y=-10x+33\)
\(y=10x-9\)
\(y=-3x+17\)
\(y=3x+5\)
Answer:

Lines are parallel if they have the same slope. The given line is in the form \(y=mx+b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. For the given line, the slope is \(m=3\), so the slope of the parallel line is also \(m=3\).

We can substitute the given point and the slope into the slope-intercept form of a line to find the equation for the line containing the point \((2,11)\). Substitute the values into the slope-intercept form, then solve for \(b\).

\(y=mx+b\)
\(11=3(2)+b\)
\(11=6+b\)
\(11-6=6+b-6\)
\(5=b\)

Plug \(b=5\) and \(m=3\) into the slope-intercept form of a line, the equation of the line containing the point \((2,11)\) is:

\(y=mx+b\)
\(y=(3)x+(5)\)
\(y=3x+5\)

Question #3:

 
Find the equation of the line that contains the point \((7,-4)\) and is parallel to the line \(y=-3x+5\).

\(y=3x+17\)
\(y=-3x+17\)
\(y=3x-19\)
\(y=-3x-5\)
Answer:

Lines are parallel when they have the same slope. The given line is in the form \(y=mx+b\), where \(m\) is the slope and \(b\) is the \(y\)-intercept. For the given line, the slope is \(m=-3\), so the slope of the parallel line is also \(m=-3\).

We can substitute the given point and the slope into the point-slope form of a line to find the equation for the line containing the point \((7,-4)\). The point-slope form of a line is:

\(y-y_1=m(x-x_1)\)

where \(m\) is the slope of the line and \((x_1,y_1)\) is a point the line passes through. Substitute the values into the point-slope form and simplify.

\(y-(-4)=-3(x-7)\)
\(y+4=-3(x-7)\)

To convert to slope-intercept form, solve for \(y\).

\(y+4=-3(x)-3(-7)\)
\(y+4=-3x+21\)

Subtract 4 from both sides of the equation.

\(y+4-4=-3x+21-4\)
\(y=-3x+17\)

Alternatively, we can substitute \(m=-3\), \(x=7\), and \(y=-4\) into the slope-intercept form, then solve for \(b\).

\(y=mx+b\)
\(-4=-3(7)+b\)
\(-4=-21+b\)
\(-4+21=-21+b+21\)
\(17=b\)

Plug \(b=17\) and \(m=-3\) into the slope-intercept form of a line, and the equation of our line containing the point \((7,-4)\) is the same.

\(y=mx+b\)
\(y=-3x+17\)

Question #4:

 
It costs Company A $8 to make a flash drive. There is a fixed cost of $100 to make the flash drives. Let \(x\) represent the number of flash drives the company makes, and \(y\) represent the total cost, in dollars, of making the flash drives. Then, the total cost of producing flash drives for Company A can be represented by the linear equation \(y=8x+100\). It also costs Company B $8 to make the same type of flash drives company A makes. If the point \((5,120)\) lies on the line that represents the total cost for Company B, which of the following is the equation of the total cost for Company B?

\(y=8x-380\)
\(y=8x+380\)
\(y=8x+80\)
\(y=8x-80\)
Answer:

Since it costs Company B the same price to make a flash drive as it does for Company A, the slope for the equation of the line for the total cost of Company B is also \(m=8\), making the two lines parallel.

We can write the equation for the total cost for Company B using the slope-intercept form \(y=mx+b\), where m is the slope and \(b\) is the \(y\)-intercept. Using the point \((5,120)\) and the slope of \(m=8\), we have:

\(y=mx+b\)
\(120=8(5)+b\)
\(120=40+b\)
\(120-40=40+b-40\)
\(80=b\)

Plug \(b=80\) and \(m=8\) into the slope-intercept form of a line, the equation of the total cost for Company B is:

\(y=mx+b\)
\(y=(8)x+(80)\)
\(y=8x+80\)

Question #5:

 
A liquid solution is heated over time. The temperature, in degrees Celsius, of the solution as it is heated can be modeled with the linear equation \(y=5x+20\), where \(x\) is time in minutes, and y is the temperature of the solution at time \(x\). The linear equation of the temperature of a second liquid solution that is heated over time is parallel to the linear equation of the temperature of the first liquid solution. If after 6 minutes the temperature was 68 degrees, which of the following is the equation of the line representing the temperature of the second solution?

\(y=5x+38\)
\(y=-5x+38\)
\(y=5x+98\)
\(y=-5x+98\)
Answer:

Lines are parallel if they have the same slope. The given line is in the form \(y=mx+b\), where m is the slope and \(b\) is the \(y\)-intercept. For the equation of the line representing the temperature of the first liquid temperature, the slope is \(m=5\), so the slope of the equation for the temperature of the second liquid solution is also \(m=5\).

We can substitute the given point and the slope into the slope-intercept of a line to find the equation for the line containing the point \((6,68)\). Substitute the values into the slope-intercept form, then solve for \(b\).

\(y=mx+b\)
\(68=5(6)+b\)
\(68=30+b\)
\(68-30=30+b-30\)
\(38=b\)

Plug \(b=38\) and \(m=5\) into the slope-intercept form of a line, the equation of the line containing the point \((6,68)\) is:

\(y=mx+b\)
\(y=(5)x+(38)\)
\(y=5x+38\)

So, the equation of the line representing the temperature of the second liquid solution is \(y=5x+38\).

 

Return to Algebra I Videos

573410

 

by Mometrix Test Preparation | This Page Last Updated: May 27, 2022