Finding the Standard Form Equation of a Line

Hello! Welcome to this video on finding the standard form equation of a line when you are given a point on the line and its slope and when you are given two points on the line. Let’s get started with an example.

Find the standard form equation for a line that has a slope of 4 and passes through the point \((6,2)\).

There are two ways to solve these types of problems. You can first solve it for slope-intercept form and then convert to standard form, or you can solve it for the point-slope form of the line and then convert that to standard form. Either way will get you the correct answer. For this first example, we are going to first solve for slope-intercept form. Remember, slope-intercept form is:

\(y=mx+b\)

Since we know the slope is 4, which is the value for \(m\), plug in 4 for \(m\) into this equation.

\(y=4x+b\)

Now, take the point you are given and plug in the values for \(x\) and \(y\) to solve for the \(y\)-intercept \((b)\).

\(2=4(6)+b\)

Now we want to solve for \(b\), so we’ll do that by first multiplying on the right side.

\(2=24+b\)

Then, subtract 24 from both sides of the equation.

\(2-24=24-24+b\)
 
\(-22=b\)

Now that we know our \(y\)-intercept, plug this into the equation from a few steps earlier. So our new equation is:

\(y=4x-22\)

Remember, this is the slope-intercept equation of our line, but we are looking for the standard form equation of our line. Standard form of a linear equation is:

\(Ax+By=C\)

So we need to rearrange to get our \(x\)-terms and our \(y\)-terms on the left side of our equation and our constants on our right side of our equation. Do this by subtracting \(4x\) from both sides of the equation.

\(y-4x=4x-4x-22\)
 
\(y-4x=-22\)

From here, switch the \(x\)– and \(y\)-terms on the left using the commutative property of addition.

\(-4x+y=-22\)

Now, remember, in order for a linear equation to be in standard form, \(A\) must be a positive integer. So, in this case, we have \(A\) as \(-4\), so we need to get rid of that negative sign. So to do that, we’re going to multiply the entire equation by \(-1\). When we do that, we’ll get:

\(-1(-4x+y=-22)\)
 
\(4x-y=22\)

And there you have it! That’s the standard form equation for our line. Let’s try another one.

Find the standard form equation of a line that has a slope of \(-2\) and passes through the point \((4,-3)\).

This time we are going to solve by using the point-slope equation of a line.

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

We know the slope \((m)\) and a point \((x_{1},y_{1})\), so plug in all the values we are given into this equation.

\(y-(-3)=-2(x-4)\)

From here, we need to convert the equation to standard form. Do this by simplifying the subtracting a negative number on the left side and distributing through the parentheses on the right side.

\(y+3=-2x+8\)

Then, subtract 3 from both sides of the equation and add \(2x\) to both sides of the equation to get the variable terms on the left side and the constants on the right side. So again, we’re going to subtract 3 from both sides.

\(y+3-3=-2x+8-3\)
\(y=-2x+5\)

And we’re going to add \(2x\) to both sides.

\(y+2x=-2x+2x+5\)
 
\(y+2x=5\)

Finally, all we have to do is switch our two terms on the left side. So this will become:

\(2x+y=5\)

Remember, we can do that because of the commutative property of addition. Now we need to check and make sure that \(A\) is a positive integer. It is, so this is our final answer.

Now let’s shift to looking at how to find the standard form equation of a line when we are given two points on the line. Notice we aren’t given the slope of this line, so in order to solve our problems, we will need to first find the slope of the line using the slope formula. Let’s try a problem.

Find the standard form equation of a line that passes through the points \((-6,1)\) and \((4,7)\).

First, find the slope of the line. Remember, the slope formula is:

\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)

If we plug in our two points and simplify, we’ll get:

\(m=\frac{7-1}{4-(-6)}=\frac{7-1}{4+6}=\frac{6}{10}=\frac{3}{5}\)

For this example, we will start by using the slope-intercept form of a line.

\(y=mx+b\)

We found the value for \(m\), so we can plug this into our equation.

\(y=\frac{3}{5}x+b\)

Now, plug in either one of the points for \(x\) and \(y\) so we can solve for \(b\). For this example, let’s use the point \((-6,1)\).

\(1=\frac{3}{5}(-6)+b\)

We’re going to start by multiplying on the right side.

\(1=-\frac{18}{5}+b\)

And then to get rid of this \(\frac{-18}{5}\), we’re going to have to add \(\frac{18}{5}\) to both sides of our equation.

\(1+\frac{18}{5}=-\frac{18}{5}+\frac{18}{5}+b\)

To be able to add 1 and \(\frac{18}{5}\), we first need to convert 1 to an equivalent fraction with a denominator of 5. So remember, any whole number is the same as the fraction of that whole number over 1. So \(\frac{1}{1}\) is the same thing as 1. So to convert this to a fraction with the denominator of 5, we’ll need to multiply both the numerator and the denominator by 5. If we do that, we’ll get:

\(\frac{5}{5}+\frac{18}{5}=b\)

Then, we can simplify that to get:

\(b=\frac{23}{5}\)

From here, let’s go back to our equation from earlier and plug in our newly found value for \(b\).

\(y=\frac{3}{5}x+\frac{23}{5}\)

From here, we need to eliminate our fractions because \(A\) will need to be an integer. To eliminate our fractions, simply multiply the entire equation by 5.

\(5(y=\frac{3}{5}x+\frac{23}{5})\)
 
\(5y=3x+23\)

Now we can subtract \(3x\) from both sides to get the variable terms on the left side and the constant on the right side.

\(5y-3x=3x-3x+23\)
 
\(5y-3x=23\)

Then, we can flip the two terms on the left side using the commutative property.

\(-3x+5y=23\)

Finally, multiply the entire equation by \(-1\) since \(A\) must be positive.

\(-1(-3x+5y=23)\)
 
\(3x-5y=-23\)

And there’s our equation! Let’s work through one more problem, this time using the point-slope equation.

Find the standard form equation of a line that passes through the points \((7,-2)\) and \((5,9)\).

The first thing we need to do is find the slope of our line using the slope formula.

\(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{9-(-2)}{5-7}=\frac{9+2}{5-7}=\frac{11}{-2}\)

Our slope is \(\frac{-11}{2}\).

Now that we know our slope, we can plug this and one of our points into the point-slope equation.

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

For this example, let’s use the point \((7,-2)\).

\(y-(-2)=-\frac{11}{2}(x-7)\)

We now have the point-slope equation of our line, so all we need to do is simplify and rearrange a bit to get our standard form equation. First, simplify the subtracting a negative number on the left side and distribute on the right side.

\(y+2=-\frac{11}{2}x+\frac{77}{2}\)

From here, we can get rid of our fractions by multiplying the entire equation by 2. This can happen in future steps as well, but doing it right here will keep us from having to subtract fractions.

\(2(y+2=-\frac{11}{2}x+\frac{77}{2})\)
 
\(2y+4=-11x+77\)

Then, subtract 4 from both sides and add \(11x\) to both sides to get the variable terms on the left side of the equation and the constants on the right side. So let’s do that all in one step. We’re going to subtract 4, and we’re going to add \(11x\). If you want to split it up into two separate steps so you don’t get confused, that’s totally fine. I just combined it in one step to show you that you can save a little bit of work while doing the same process.

\(2y+4-4+11x=-11x+11x+77-4\)
\(2y+11x=73\)

Last but not least, we have to flip the two terms on the left side of the equation using the commutative property.

\(11x+2y=73\)

And that’s our answer!

I hope this video on finding the standard form equation of a line was helpful. Thanks for watching, and happy studying!

Standard Form Equation of a Line Practice Questions