7.5: Standard Form of a Line
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In this section we will investigate the standard form of a line. Let’s begin with a simple example.
Example 7.5.1
Solve the equation 2x+3y=6 for y and plot the result.
Solution
First we solve the equation 2x+3y=6 for y. Begin by isolating all terms containing y on one side of the equation, moving or keeping all the remaining terms on the other side of the equation.
2x+3y=6 Original equation. 2x+3y−2x=6−2x Subtract 2x from both sides. 3y=6−2x Simplify. 3y3=6−2x3 Divide both sides by 3
Note
Just as multiplication is distributive with respect to addition a(b+c)=ab+acso too is division distributive with respect to addition.a+bc=ac+bc
When dividing a sum or a difference by a number, we use the distributive property and divide both terms by that number.
y=63−2x3 On the left, simplify. On the right, divide both terms by 3y=2−2x3 Simplify.
Finally, use the commutative property to switch the order of the terms on the right-hand side of the last result.
y=2+(−2x3) Add the opposite. y=−23x+2 Use the commutative property to switch the order.
Because the equation 2x+3y=6 is equivalent to the equation y=−23x+2, the graph of 2x+3y=6 is a line, having slope m=−2/3 and y-intercept (0,2). Therefore, to draw the graph of 2x+3y=6, plot they-intercept (0,2), move down 2 and 3 to the right, then draw the line (see Figure 7.5.1 ).

Exercise 7.5.1
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- Answer
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In general, unless B=0, we can always solve the equation Ax+By=C for y:
Ax+By=C Original equation. Ax+By−Ax=C−Ax Subtract Ax from both sides. By=C−Ax Simplify. ByB=C−AxB Divide both sides by By=CB−AxB distribute the By=−ABx+CB Commutative property
Note that the last result is in slope-intercept form y=mx+b, whose graph is a line. We have established the following result.
FAct
The graph of the equation Ax+By=C, is a line.
Important points: A couple of important comments are in order.
- The form Ax+By=C requires that the coefficients A, B, and C are integers. So, for example, we would clear the fractions from the form 12x+23y=14by multiplying both sides by the least common denominator. 12(12x+23y)=(14)126x+8y=3Note that the coefficients are now integers.
- The form Ax+By=C also requires that the first coefficient A is nonnegative; i.e., A≥0. Thus, if we have −5x+2y=6then we would multiply both sides by −1, arriving at: −1(−5x+2y)=(6)(−1)5x−2y=−6Note that A=5 is now greater than or equal to zero.
- If A, B, and C have a common divisor greater than 1, it is recommended that we divide both sides by the common divisor, thus “reducing” the coefficients. For example, if we have 3x+12y=−24then dividing both side by 3 “reduces” the size of the coefficients.3x+12y3=−243x+4y=−8
Standard form
The form Ax+By=C, where A, B, and C are integers, and A≥0, is called the standard form of a line.
Slope-Intercept to Standard Form
We’ve already transformed a couple of equations in standard form into slopeintercept form. Let’s reverse the process and place an equation in slope intercept form into standard form.
Example 7.5.2
Given the graph of the line in Figure 7.5.2, find the equation Given the graph of the line below, find the equation of the line in standard form.

Solution
The line intercepts the y-axis at (0,−3). From (0,−3), move up 5 units, then left 2 units. Thus, the line has slope Δy/Δx=−5/2 (see Figure 7.5.3). Substitute −5/2 form and −3 for b in the slope-intercept form of the line.

y=mx+b Slope-intercept form. y=−52x−3 Substitute: −5/2 for m,−3 for b
To put this result in standard form Ax+By=C, first clear the fractions by multiplying both sides by the common denominator.
2y=2[−52x−3] Multiply both sides by 22y=2[−52x]−2[3] Distribute the 22y=−5x−6 Multiply.
That clears the fractions. To put this last result in the form Ax+By=C, we need to move the term −5x to the other side of the equation.
5x+2y=−5x−6+5x Add 5x to both sides. 5x+2y=−6 Simplify.
Thus, the standard form of the line is 5x+2y=−6. Note that all the coefficients are integers and the terms are arranged in the order Ax+By=C, with A≥0.
Exercise 7.5.2
Given the graph of the line below, find the equation of the line in standard form.
- Answer
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3x−4y=−2
Point-Slope to Standard Form
Let’s do an example where we have to put the point-slope form of a line in standard form.
Example 7.5.3
Sketch the line passing through the points (−3,−4) and (1,2), then find the equation of the line in standard form.
Solution
Plot the points (−3,−4) and (1,2), then draw a line through them (see Figure 7.5.4).

Use the points (−3,−4) and (1,2) to calculate the slope.
Slope =ΔyΔx Slope formula. =2−(−4)1−(−3) Subtract coordinates of (−3,−4)=64 Simplify. =32 Reduce.
Let’s substitute (x0,y0)=(1,2) and m=3/2 in the point-slope form of the line. (Note: Substituting (x0,y0)=(−3,−4) and m=3/2 would yield the same answer.)
y−y0=m(x−x0) Point-slope form. y−2=32(x−1) Substitute: 3/2 for m,1 for x0
The question requests that our final answer be presented in standard form. First we clear the fractions.
y−2=32x−32 Distribute the 3/22[y−2]=2[32x−32] Multiply both sides by 22y−2[2]=2[32x]−2[32] Distribute the 22y−4=3x−3 Multiply.
Now that we’ve cleared the fractions, we must order the terms in the form Ax+By=C. We need to move the term 3x to the other side of the equation.
2y−4−3x=3x−3−3x Subtract 3x from both sides. −3x+2y−4=−3 Simplify, changing the order on the left-hand side.
To put this in the form Ax+By=C, we need to move the term −4 to the other side of the equation.
−3x+2y−4+4=−3+4 Add 4 to both sides. −3x+2y=1 Simplify.
It appears that −3x+2y=1 is in the form Ax+By=C. However, standard form requires that A≥0. We have A=−3. To fix this, we multiply both sides by −1.
−1[−3x+2y]=−1[1] Multiply both sides by −13x−2y=−1 Distribute the −1
Thus, the equation of the line in standard form is 3x−2y=−1.
Note
If we fail to reduce the slope to lowest terms, then the equation of the line would be: y−2=64(x−1)
Multiplying both sides by 4 would give us the result 4y−8=6x−6or equivalently: −6x+4y=2
This doesn’t look like the same answer, but if we divide both sides by −2, we do get the same result. 3x−2y=−1
This shows the importance of requiring A≥0 and “reducing” the coefficients A, B, and C. It allows us to compare our answer with our colleagues or the answers presented in this textbook.
Exercise 7.5.3
Find the standard form of the equation of the line that passes through the points (−2,4) and (3,−3).
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7x+5y=6
Intercepts
We’ve studied the y-intercept, the point where the graph crosses the y-axis, but equally important are the x-intercepts, the points where the graph crosses the x-axis.

In Figure 7.5.5, the graph crosses the x-axis three times. Each of these crossing points is called an x-intercept. Note that each of these x-intercepts has a y-coordinate equal to zero. This leads to the following rule.
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To find the x-intercepts of the graph of an equation, substitute y=0 into the equation and solve for x.

Similarly, the graph in Figure 7.5.6 crosses the y-axis three times. Each of these crossing points is called a y-intercept. Note that each of these y-intercepts has an x-coordinate equal to zero. This leads to the following rule.
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To find the y-intercepts of the graph of an equation, substitute x=0 into the equation and solve for y.
Let’s put these rules for finding intercepts to work.
Example 7.5.4
Find the x- and y-intercepts of the line having equation 2x−3y=6. Plot the intercepts and draw the line.
Solution
We know that the graph of 2x−3y=6 is a line. Furthermore, two points completely determine a line. This means that we need only plot the x- and y-intercepts, then draw a line through them.
To find the x-intercept of 2x−3y=6, substitute 0 for y and solve for x.
2x−3y=62x−3(0)=62x=62x2=62x=3
Thus, the x-intercept of the line is (3,0).
To find the y-intercept of 2x−3y=6, substitute 0 for x and solve for y.
2x−3y=62(0)−3y=6−3y=6−3y−3=6−3y=−2
Thus, the y-intercept of the line is (0,−2).
Plot the x-intercept (3,0) and the y-intercept (0,−2) and draw a line through them (see Figure 7.5.7).

Exercise 7.5.4
Find the x- and y-intercepts of the line having equation 3x+4y=−12. Plot the intercepts and draw the line.
- Answer
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x-intercept: (−4,0)
y-intercept: (0,−3)
Example 7.5.5
Sketch the line 4x+3y=12, then sketch the line through the point (−2,−2) that is perpendicular to the line 4x+3y=12. Find the equation of this perpendicular line.
Solution
Let’s first find the x- and y-intercepts of the line 4x+3y=12.
To find the x-intercept of the line 4x+3y=12, substitute 0 for y and solve for x.
4x+3y=124x+3(0)=124x=124x4=124x=3
Thus, the x-intercept of the line is (3,0).
To find the y-intercept of the line 4x+3y=12, substitute 0 for x and solve for y.
4x+3y=124(0)+3y=123y=123y3=123y=4
Thus, the y-intercept of the line is (0,4).
Plot the intercepts and draw a line through them. Note that it is clear from the graph that the slope of the line 3x+4y=12 is −4/3 (see Figure 7.5.8).

You could also solve for y to put 3x+4y=12 in slope intercept form in order to determine the slope.
Because the slope of 3x+4y=12 is −4/3, the slope of a line perpendicular to 3x+4y=12 will be the negative reciprocal of −4/3, namely 3/4. Our perpendicular line has to pass through the point (−2,−2). Start at (−2,−2), move 3 units upward, then 4 units to the right, then draw the line. It should appear to be perpendicular to the line 3x+4y=12 (see Figure 7.5.9).

Finally, use the point-slope form, m=3/4, and (x0,y0)=(−2,−2) to determine the equation of the perpendicular line.
y−y0=m(x−x0) Point-slope form. y−(−2)=34(x−(−2)) Substitute: 3/4 for m,−2 for x0 and −2 for y0y+2=34(x+2) Simplify.
Let’s place our answer in standard form. Clear the fractions.
y+2=34x+64 Distribute 3/44[y+2]=4[34x+64] Multiply both sides by 44y+4[2]=4[34x]+4[64] Distribute the 44y+8=3x+6 Multiply.
Rearrange the terms to put them in the order Ax+By=C.
4y+8−3x=3x+6−3x Subtract 3x from both sides. −3x+4y+8=6 Simplify. Rearrange on the left. −3x+4y+8−8=6−8 Subtract 8 from both sides. −3x+4y=−2 Simplify. −1(−3x+4y)=−1(−2) Multiply both sides by −13x−4y=2 Distribute the −1
Hence, the equation of the perpendicular line is 3x−4y=2.
Exercise 7.5.5
Find the equation of the line that passes through the point (3,2) and is perpendicular to the line 6x−5y=15.
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5x+6y=27
Horizontal and Vertical Lines
Here we keep an earlier promise to address what happens to the standard form Ax+By=C when either A=0 or B=0. For example, the form 3x=6, when compared with the standard form Ax+By=C, has B=0. Similarly, the form 2y=−12, when compared with the standard form Ax+By=C, has A=0. Of course, 3x=6 can be simplified to x=2 and 2y=−12 can be simplified to y=−6. Thus, if either A=0 or B=0, the standard form Ax + By = C takes the form x=a and y=b, respectively.
As we will see in the next example, the form x=a produces a vertical line, while the form y=b produces a horizontal line.
Example 7.5.6
Sketch the graphs of x=3 and y=−3.
Solution
To sketch the graph of x=3, recall that the graph of an equation is the set of all points that satisfy the equation. Hence, to draw the graph of x=3, we must plot all of the points that satisfy the equation x=3; that is, we must plot all of the points that have an x-coordinate equal to 3. The result is shown in Figure 7.5.10.

Secondly, to sketch the graph of y=−3, we plot all points having a y-coordinate equal to −3. The result is shown in Figure 7.5.11.

Things to note:
A couple of comments are in order regarding the lines in Figures 7.5.10 and 7.5.11.
- The graph of x=3 in Figure 7.5.10, being a vertical line, has undefined slope. Therefore, we cannot use either of the formulae y=mx+b or y−y0=m(x−x0) to obtain the equation of the line. The only way we can obtain the equation is to note that the line is the set of all points (x,y) whose x-coordinate equals 3.
- However, the graph of y=−3, being a horizontal line, has slope zero, so we can use the slope-intercept form to find the equation of the line. Note that the y-intercept of this graph is (0,−3). If we substitute these numbers into y=mx+b, we get:y=mx+b Slope-intercept form. y=0x+(−3) Substitute: 0 for m,−3 for by=−3 Simplify.
However, it is far easier to just look at the line in Figures 7.5.11 and note that it is the collection of all points (x,y) with y=3.
Exercise 7.5.6
Sketch the graphs of x=−2 and y=2.
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