# 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 $$y=2x+1$$.

$$y=-2x+1$$
$$y=2x-11$$
$$y=x-8$$
$$y=-x+1$$

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 $$y=3x-10$$.

$$y=-10x+33$$
$$y=10x-9$$
$$y=-3x+17$$
$$y=3x+5$$

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 $$y=-3x+5$$.

$$y=3x+17$$
$$y=-3x+17$$
$$y=3x-19$$
$$y=-3x-5$$

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?

$$y=8x-380$$
$$y=8x+380$$
$$y=8x+80$$
$$y=8x-80$$

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?

$$y=5x+38$$
$$y=-5x+38$$
$$y=5x+98$$
$$y=-5x+98$$

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 $$y=5x+38$$.