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
Write the equation of a line in standard form that has a slope of 3 and contains the point \((4,-2)\).
\(3x-y=-10\)
\(3x+y=10\)
\(3x-y=14\)
\(3x+y=-14\)
The equation of a line in standard form is:
\(Ax+By=C\)
where \(A\), \(B\), and \(C\) are integers, \(A>0\), and both \(A\) and \(B\) are not zero.
We can write the equation of a line using the slope-intercept form or the point-slope form, then convert it to standard form. We will use the slope-intercept form, \(y=mx+b\), where \(m\) is the slope of the line, and \(b\) is the \(y\)-intercept. First, substitute \(m=3\) into the equation.
\(y=3x+b\)
Next, substitute the coordinates of the given point, \(x=4\) and \(y=-2\) into the equation and solve for \(b\).
\(-2=3\left(4\right)+b\)
\(-2=12+b\)
\(-14=b\)
Now, substitute the value of \(b\) into the equation to write the linear equation of the line in slope-intercept form.
\(y=3x-14\)
To put the equation in standard form, put the \(x\)- and \(y\)-terms on the same side and make the coefficient for the \(x\)-term positive.
\(y=3x-14\)
\(-3x+y=-14\)
\(–1-3x+y=-14\)
\(3x-y=14\)
Write the equation of a line in standard form that contains the points \((3,1)\) and \((1,9)\).
\(4x+y=13\)
\(4x-y=-13\)
\(4x+y=7\)
\(4x-y=-7\)
The equation of a line in standard form is:
\(Ax+By=C\)
where \(A\), \(B\), and \(C\) are integers, \(A>0\), and both \(A\) and \(B\) are not zero.
We can write the equation of a line using either the slope-intercept form or the point-slope form, then convert it to standard form. We will use the point-slope form of a line,
\(y-y_1=m(x-x_1)\)
where m is the slope of the line, \((x_1,y_1)\) is a point that the line passes through, and \((x,y)\) is any point on the line.
First, we need to find the slope of the line using the given points. Use the slope formula, \(m=\frac{y_2-y_1}{x_2-x_1}\), where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute the known information and simplify to find m.
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{9-1}{1-3}=\frac{\ 8}{-2}=-4\)
Next, substitute the slope of \(m=-4\) into the point-slope formula.
\(y-y_1=-4(x-x_1)\)
Now, substitute either point into the formula for \(x_1\) and \(y_1\) to put the equation of the line in standard form. Substitute the point \((3,1)\) and simplify.
\(y-\left(1\right)=-4\left(x-3\right)\)
\(y-1=-4x+12\)
To put the equation in standard form, put the \(x\)- and \(y\)-terms on the same side and make the coefficient for the \(x\)-term positive.
\(y-1=-4x+12\)
\(y-1+1=-4x+12+1\)
\(y+4x=-4x+13+4x\)
\(4x+y=13\)
Write the equation of a line in standard form that contains the points \((2,-3)\) and \((5,1)\).
\(4x+3y=-17\)
\(4x-3y=17\)
\(4x-3y=-18\)
\(4x+3y=18\)
The equation of a line in standard form is \(Ax+By=C\), where \(A\), \(B\), and \(C\) are integers, \(A>0\), and both \(A\) and \(B\) are not zero.
We can write the equation of a line using either the slope-intercept form or the point-slope form, then convert it to standard form. We will use the point-slope form of a line, \(y-y_1=m(x-x_1)\), where m is the slope of the line, \((x_1,y_1)\) is a point that the line passes through, and \((x,y)\) is any point on the line.
First, we need to find the slope of the line using the given points. Use the slope formula, \(m=\frac{y_2-y_1}{x_2-x_1}\), where \((x_1,y_1)\) and \((x_2,y_2)\) are two points on the line. Substitute our known information and solve for \(m\).
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-(-3)}{5-2}=\frac{1+3}{3}=\frac{4}{3}\)
Next, substitute the slope of \(m=\frac{4}{3}\) into the point-slope formula.
\(y-y_1=\frac{4}{3}(x-x_1)\)
Now, substitute either point into the formula for \(x_1\) and \(y_1\) to put the equation of the line in point-slope form. Substitute the point \((2,-3)\) and simplify.
\(y-\left(-3\right)=\frac{4}{3}\left(x-2\right)\)
\(y+3=\frac{4}{3}x-\frac{4}{3}\left(2\right)\)
\(y+3=\frac{4}{3}x-\frac{8}{3}\)
Before putting the equation in standard form, get rid of the fractions by multiplying both sides by 3.
\(3\left(y+3=\frac{4}{3}x-\frac{8}{3}\right)\)
\(3y+9=4x-8\)
Now, put the equation in standard form by putting the \(x\)- and \(y\)-terms on the same side making sure the coefficient for the \(x\)-term is positive.
\(3y+9-9=4x-8-9\)
\(3y=4x-17\)
\(3y-4x=4x-17-4x\)
\(-4x+3y=-17\)
\(-1\left(-4x+3y=-17\right)\)
\(4x-3y=17\)
You are training to reduce the time it takes you to run 2 miles. The first day it takes you 25 minutes to run 2 miles, and on the eleventh day it takes you only 20 minutes to run the same distance. Suppose the time it takes you to run 2 miles can be modeled by a linear equation where \(y\) is the time it takes you to run 2 miles, and \(x\) is the day you run it. Which of the following is the linear equation in standard form that represents the time it took you to run the distance on a given day?
\(x-2y=-23\)
\(x+2y=23\)
\(x-2y=-51\)
\(x+2y=51\)
Since \(x\) represents a day, and \(y\) is the time it takes you to run 2 miles on that day, we can write the ordered pairs \((1,25)\) and \((11,20)\) to represent the day and time it takes to run 2 miles, respectively.
The equation of a line in standard form is \(Ax+By=C\), where \(A\), \(B\), and \(C\) are integers, \(A>0\), and both \(A\) and \(B\) are not zero.
We can write the equation of a line using either the slope-intercept form or the point-slope form, then convert it to standard form. We will use slope-intercept form, \(y=mx+b\), where \(m\) is the slope of the line, and \(b\) is the \(y\)-intercept.
First, we need to 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 our known information and simplify to find \(m\).
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{20-25}{11-1}=-\frac{5}{10}=-\frac{1}{2}\)
Now, substitute \(m=-\frac{1}{2}\) into the equation.
\(y=-\frac{1}{2}x+b\)
Next, substitute the coordinates of the one of the points into the equation and solve for \(b\). Substitute \(\left(1,25\right)\).
\(25=-\frac{1}{2}\left(1\right)+b\)
\(25=-\frac{1}{2}+b\)
\(25+\frac{1}{2}=-\frac{1}{2}+b+\frac{1}{2}\)
\(\frac{51}{2}=b\)
Now, substitute the value of \(b\) into the equation to write the linear equation of the line in slope-intercept form.
\(y=-\frac{1}{2}x+\frac{51}{2}\)
Before putting the equation in standard form, get rid of the fractions by multiplying both sides by 2.
\(2\left(y=-\frac{1}{2}x+\frac{51}{2}\right)\)
\(2y=-x+51\)
Now, put the equation in standard form by putting the \(x\)- and \(y\)-terms on the same side and make the coefficient for the \(x\)-term positive if needed.
\(2y+x=-x+51+x\)
\(x+2y=51\)
A homeowner is charged a fixed cost plus an additional cost per hour to have a painter paint his home. The painter charges $80 for two hours of painting the home, and $240 for ten hours of painting the home. The cost of painting the home can be modeled with a linear equation where \(y\) is the cost and \(x\) is the number of hours painted. Which of the following is a linear equation in standard form that represents this situation?f
\(20x+y=40\)
\(20x-y=-40\)
\(2x-y=470\)
\(2x+y=-470\)
Since \(x\) represents the number of hours the painter paints, and \(y\) is the cost the painter charges for working \(x\) hours, we can write the ordered pairs \((2,80)\) and \((10,240)\) to represent number of hours the painter painted and the cost charged, respectively.
The equation of a line in standard form is \(Ax+By=C\), where \(A\), \(B\), and \(C\) are integers, \(A>0\), and both \(A\) and \(B\) are not zero.
We can write the equation of a line using either the slope-intercept form or the point-slope form, then convert it to standard form. We will use the point-slope form of a line, \(y-y_1=m(x-x_1)\), where \(m\) is the slope of the line, and \((x_1,y_1)\) is a point the line passes through.
First, we need to 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. Substituting our known information and solving for \(m\) gives us:
\(m=\frac{y_2-y_1}{x_2-x_1}=\frac{240-80}{10-2}=\frac{160}{8}=20\)
Next, substitute the slope of \(m=20\) into the point-slope formula.
\(y-y_1=20(x-x_1)\)
Now, substitute either point into the formula for \(x_1\) and \(y_1\) to put the equation of the line in point-slope form. Substitute the point \((10,240)\).
\(y-240=20\left(x-10\right)\)
\(y-240=20x-200\)
Lastly, put the equation in standard form by putting the \(x\)- and \(y\)-terms on the same side making sure the coefficient for the \(x\)-term is positive.
\(y-240+240=20x-200+240\)
\(y-20x=20x+40-20x\)
\(-20x+y=40\)
\(-1(-20x+y=40)\)
\(20x-y=-40\)
