\(Ax+By+C=0\)


Explore This 1


Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/mqtuz7w4#


Explore This Summary

In the Explore This, you may have noticed the following:

  • When performing an operation on a linear relation, the equation of the line looks different.
  • Graphing the equation of a line after completing the operation shows that, although the equation looks different, it represents the same line.

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Equations of Lines

We know that all points that lie on a line have an \(x\)- and \(y\)-coordinate that will satisfy the equation. 

 

\(Ax+By+C=0\)​​​​​

A linear relation can take the form \(Ax+By+C=0\), where

 

 

Example 1

Express the equation of a line given as \(3x-5y+1=0\) in the form \(y=mx+b\).

 

 

Example 1 Continued

Express the equation of a line given as \(3x-5y+1=0\) in the form \(y=mx+b\).

Solution

Graphing \(3x-5y+1=0\) and \(y=\dfrac{3}{5}x+\dfrac{1}{5}\), 

\(Ax+By+C=0\) Continued

In the previous example, we saw that \(3x-5y+1=0\) and \(y=\dfrac{3}{5}x+\dfrac{1}{5}\) are equivalent.

       

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      Example 2

      Given  \(y=\dfrac{2}{3}x+7\), write the equation of the line in the form \(Ax+By+C=0.\)

      Solution

      We would like to write the equation in the form \(Ax+By+C=0\), where \(A\), \(B\), and \(C\) are integers, and \(A\) is positive.

      \(y\)

       \(=\dfrac{2}{3}x+7\)

       

      \(3y \)

       ​​\(=2x+21\)

      • Multiply each term by \(3\) to eliminate fractions.

      \(-2x+3y-21\)

       \(=0\)

      • Use inverse operations to move every term to one side.

      \(2x-3y+21\)

       \(=0\)

      • Multiply every term by \(-1\) so \(A\) is a positive value.

      To avoid needing to multiply each term by \(-1\), we could have moved \(3y\) to the right side of the equation to get \(0=2x-3y+21\) as shown here.

      \(y\)

       \(=\dfrac{2}{3}x+7\)

       

      \(3y\)

       \(=2x+21\)

      • Multiply each term by \(3\) to eliminate fractions.

      \(0\)

       \(=2x-3y+21\)

      • Use inverse operations to move every term to one side.

      Check Your Understanding 1


      Given \(y = (((((((((((((((((s)*(l))*(o))*(p))*(e))*(I))*(n))*(t))*(e))*(r))*(c))*(e))*(p))*(t))*(F))*(o))*(r))*(m)\), rearrange the equation of the line into the form \(Ax+By+C=0\) where \(A\), \(B\), and \(C\) are integers, and \(A\) is positive.

      Remember to include both the left and right sides of the equation.

      There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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      How Did I Do?Try Another

      Benefits of \(Ax+By+C=0\)

      Some benefits of the form \(Ax+By+C=0\) are:

      • We can quickly identify the \(x\)-intercept by substituting in \(y=0\) and solving for \(x\), and we can identify the  \(y\)-intercept by substituting in \(x=0\) and solving for \(y\).
      • We can usually graph the line quickly using the \(x\)- and \(y\)-intercepts.

      Example 3

      Heather is baking chocolate cupcakes and making hot chocolate. The cupcake recipe calls for \(4\) tablespoons of cocoa for each batch, and the hot chocolate recipe calls for \(1\) tablespoon of cocoa for each glass. Heather has exactly \(16\) tablespoons of cocoa she wants to use up.

      1. Determine a linear equation to model this situation.
      2. Graph the equation of the line.
      3. What information does the graph provide?

      Solution — Part A

      We know that Heather can make batches of cupcakes and glasses of hot chocolate based on the amount of cocoa she has left.

      • Let \(c\) represent the number of batches of cupcakes.
      • Let \(h\) represent the number of glasses of hot chocolate.

      Because there are \(4\) tablespoons of cocoa needed for a batch of cupcakes and \(1\) tablespoon of cocoa needed for a glass of hot chocolate, we can represent this situation with the equation:

      \(4c+1h=16\)

      OR

      \(4c+h=16\)

      Notice that this equation is in the form \(Ax+By=C\). This is another common form of an equation of a line.

      Solution — Part B

      This is an example where there really isn't a clear independent or dependent variable.

      We could say:

      • The number of batches of cupcakes depends on the number of glasses of hot chocolate.

      However, we could also say:

      • The number of glasses of hot chocolate depends on the number of batches of cupcakes.

      Case 1

      Let's consider a graph of \(4c+h=16\), where the number of batches of cupcakes depends on how many glasses of hot chocolate are made.

      If we rearrange for \(c\) in terms of \(h\), we get:

      \(\begin{align*} 4c+h&=16\\ 4c&=16-h\\ c&=\dfrac{16}{4}-\dfrac{h}{4}\\ c&=-\dfrac{1}{4}h+4 \end{align*}\)

      This gives us an equation where we could say that the number of batches of cupcakes depends on the number of glasses of hot chocolate.

      Graphing  \(c=-\dfrac{1}{4}h+4\), we see:

      • a \(c\)-intercept of \(4\), and
      • an \(h\)-intercept of \(16\).

      A graph of the function c.

      Take a moment to think about what the intercept values mean for the situation described in this question.

      Case 2

      Let's consider a graph of \(4c+h=16\), where the number of glasses of hot chocolate depends on how many batches of cupcakes are made.

      If we rearrange for \(h\) in terms of \(c\), we get:

      \(\begin{align*} 4c+h&=16\\ h&=16-4c\\ h&=-4c+16\\ \end{align*}\)

      This gives us an equation where we could say that the number of glasses of hot chocolate depends on the number of batches of cupcakes.

      Graphing \(h=-4c+16\), we see:

      • an \(h\)-intercept of \(16\), and
      • a \(c\)-intercept of \(4\).

      A graph of the function h.

      Solution — Part C

      The graph of \(4c+h=16\) can help to answer the question: "What are possible pairs of values for the number of batches of cupcakes and glasses of hot chocolate that Heather can make with her \(16\) tablespoons of cocoa?"

      Every point along the line of \(4c+h=16\), represents a solution to the equation.

      Although graphing \(h\) in terms of \(c\), and then \(c\) in terms of \(h\), gave two different looking lines, both arrangements of the equation provide the same pairs of solutions; they are simply written in a different order.

      • \(c=-\dfrac{1}{4}h+4\) gives the ordered pairs \((h,c)\).
      • \(h=-4c+16\) gives the ordered pairs \((c,h)\).

      For example, the \(c\)-intercept for \(c=-\dfrac{1}{4}h+4\), is located at the point \((0,4)\). The \(c\)-intercept for \(h=-4c+16\), is located at \((4,0)\). In both cases, \(c=4\) and \(h=0\), meaning that if Heather makes \(4\) batches of cupcakes, then she can cannot make any glasses of hot chocolate.

      Does every point along the line of \(4c+h=16\) represent a possible answer for the number of batches of cupcakes and glasses of hot chocolate Heather can make with \(16\) tablespoons of cocoa?

      • We know that a line is made up of an infinite number of points.
      • Let's eliminate some possible solutions by noting that the number of batches of cupcakes and glasses of hot chocolate cannot be a negative number. We know:  \(c\ge 0\) and \(h\ge 0\).

      • If both the number of batches of cupcakes and glasses of hot chocolate are greater than or equal to \(0\), then the possible solutions are limited to the first quadrant on the graphs (or the values along the line between and including the intercepts.)

      The graph of the function c is highlighted in the first quadrant.

      The graph of the function h is highlighted in the first quadrant.

      • Therefore, \(0\le c\le4\) and \(0 \le h \le 16\), where \(c\) and \(h\) are real numbers.
      • Although real numbers are possible solutions (that is, we can have partial batches and glasses), integer solutions are more relevant in this situation and would include:

      \(c=4\),  \(h=0\)

      \(c=3\),  \(h=4\)

      \(c=2\),  \(h=8\)

      \(c=1\),  \(h=12\)

      \(c=0\),  \(h=16\)

      Note that we can get these values by using the given equation and substituting in values for \(c\) and solving for \(h\) or vice versa. This would be similar to creating a table of values.

      Benefits of \(Ax+By=C\) 

      Some benefits of the form \(Ax+By=C\) are:

      • We can determine equations from scenarios where the context does not provide clear independent and dependent variables. 
      • We can usually graph the line quickly using \(x\)- and \(y\)-intercepts, especially if \(A\) and \(B\) are factors of \(C\).

      Check Your Understanding 2


      (((n)*(a))*(m))*2.718281828459045 had two summer jobs. As a (((j)*(o))*(b))*1.0 (((n)*(a))*(m))*2.718281828459045 earned \($((((((((((((((j)*(o))*(b))*1.0)*(r))*(a))*(t))*(e))*(D))*(i))*(s))*(p))*(l))*(a))*(y)\) per hour. As a (((j)*(o))*(b))*2.0 (((n)*(a))*(m))*2.718281828459045 earned \($((((((((((((((j)*(o))*(b))*2.0)*(r))*(a))*(t))*(e))*(D))*(i))*(s))*(p))*(l))*(a))*(y)\) per hour. (((n)*(a))*(m))*2.718281828459045 earned a total of \($((((((((((t)*(o))*(t))*(a))*(ln(o)))*(C))*(o))*(m))*(m))*(a))*(s)\) over the summer.

      Write a linear equation to model this relationship, where \($letter1(details...)\) represents the number of hours worked as a (((j)*(o))*(b))*1.0 and \($letter2(details...)\) represents the number of hours worked as a (((j)*(o))*(b))*2.0 over the summer.

      Remember to include both the left and right sides of the equation. Do not include dollar signs (\($\)).

      There appears to be a syntax error in the question bank involving the question field of this question. The following error message may help correct the problem:

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      How Did I Do?Try Another

      Try This Revisited

      Do you agree or disagree with the statement: "All of the following equations represent the same line"?

      Justify your decision.

      1. \(y=\dfrac{7}{6}x-\dfrac{1}{2}\)
      2. \(7x-6y-3=0\)
      3. \(x=\dfrac{6}{7}y+\dfrac{3}{7}\)
      4. \(7x-6y=3\)

      Solution

      The equations listed all represent the same line. To show this is true, we have a couple of options.

      Approach 1

      We could input each equation into a graphing calculator and see that all of the equations produce the same line.

      The earlier equations all lay on the same line.

      Approach 2

      We could rearrange each equation to be in the same form and show they are identical. Since we have worked most frequently with the form \(y=mx+b\), we will choose this one, however, any form would work.

      1. \(y=\dfrac{7}{6}x-\dfrac{1}{2}\) does not need to be rearranged.

      2. The second equation needs to be rearranged.

        \[\begin{align*} 7x-6y-3&=0\\ 7x-6y&=3\\ -6y&=-7x+3\\ y&=\dfrac{-7}{-6}x+\dfrac{3}{-6}\\ y&=\dfrac{7}{6}x-\dfrac{1}{2} \end{align*}\]

      3. The third equation needs to be rearranged.

        \[\begin{align*} x&=\dfrac{6}{7}y+\dfrac{3}{7}\\ 7x&=6y+3\\ 7x-6y&=3\\ -6y&=-7x+3\\ y&=\dfrac{-7}{-6}x+\dfrac{3}{-6}\\ y&=\dfrac{7}{6}x-\dfrac{1}{2} \end{align*}\]

      4. The fourth equation needs to be rearranged also.

        \[\begin{align*} 7x-6y&=3\\ -6y&=-7x+3\\ y&=\dfrac{-7}{-6}x+\dfrac{3}{-6}\\ y&=\dfrac{7}{6}x-\dfrac{1}{2} \end{align*}\]

      Therefore, all four equations represent the same line.

      In the simplification above, notice that the first line of Equation 4 is equal to the third line of Equation 3, which is equal to the second line of Equation 2. If we had recognized this, we could have stopped our simplification of Equations 3 and 4 at these points and concluded that each equation was equivalent to the previous equation, and thus all four equations represent the same line.


      Check Your Understanding 3


      Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/yd2uneaq#