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 $\PageIndex{1}$

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.

\[\begin{aligned} 2x+3y &= 6 \quad \color{Red} \text { Original equation. } \\ 2x+3y-2x &= 6-2x \quad \color{Red} \text { Subtract } 2x \text { from both sides. } \\ 3y &= 6-2x \quad \color{Red} \text { Simplify. } \\ \dfrac{3y}{3} &= \dfrac{6-2x}{3} \quad \color{Red} \text { Divide both sides by } 3 \end{aligned} \nonumber \]

Note

Just as multiplication is distributive with respect to addition \[a(b + c)=ab + ac \nonumber \]so too is division distributive with respect to addition.\[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c} \nonumber \]

When dividing a sum or a diﬀerence by a number, we use the distributive property and divide both terms by that number.

\[\begin{aligned} y &= \dfrac{6}{3}-\dfrac{2x}{3} \quad \color{Red} \text { On the left, simplify. On the right, divide both terms by } 3 \\ y &= 2-\dfrac{2 x}{3} \quad \color{Red} \text { Simplify. } \end{aligned} \nonumber \]

Finally, use the commutative property to switch the order of the terms on the right-hand side of the last result.

\[\begin{aligned} y &= 2+\left(-\dfrac{2 x}{3}\right) \quad \color{Red} \text { Add the opposite. } \\ y &= -\dfrac{2}{3} x+2 \quad \color{Red} \text { Use the commutative property to switch the order. } \end{aligned} \nonumber \]

Because the equation $2x +3y = 6$ is equivalent to the equation $y=-\dfrac{2}{3} x+2$, the graph of $2x +3 y = 6$ 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 $\PageIndex{1}$ ).

fig 3.6.1.png — Figure $\PageIndex{1}$: The graph of $2x +3y = 6$, or equivalently, $y=-\dfrac{2}{3} x+2$

Exercise $\PageIndex{1}$

Add exercises text here.

Answer: $Exercise 3.6.1.png$

In general, unless $B = 0$, we can always solve the equation $Ax + By = C$ for y:

\[\begin{aligned}
Ax+By &= C \quad \color{Red} \text { Original equation. } \\
Ax+By-Ax &= C-Ax \quad \color{Red} \text { Subtract } Ax \text { from both sides. } \\
By &= C-Ax \quad \color{Red} \text { Simplify. } \\
\dfrac{By}{B} &= \dfrac{C-Ax}{B} \quad \color{Red} \text { Divide both sides by } B \\
y &= \dfrac{C}{B}-\dfrac{Ax}{B} \quad \color{Red} \text { distribute the } B \\
y &= -\dfrac{A}{B}x+\dfrac{C}{B} \quad \color{Red} \text { Commutative property }
\end{aligned} \nonumber \]

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 \[\dfrac{1}{2} x+\dfrac{2}{3} y=\dfrac{1}{4} \nonumber \]by multiplying both sides by the least common denominator. \[\begin{aligned} 12\left(\dfrac{1}{2} x+\dfrac{2}{3} y\right) &=\left(\dfrac{1}{4}\right) 12 \\ 6 x+8 y &=3 \end{aligned}\nonumber \]Note 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 \[-5 x+2 y=6 \nonumber \]then we would multiply both sides by $−1$, arriving at: \[\begin{aligned}-1(-5 x+2 y) &=(6)(-1) \\ 5 x-2 y &=-6 \end{aligned} \nonumber \]Note 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 \[3x + 12y =−24 \nonumber \]then dividing both side by $3$ “reduces” the size of the coeﬃcients.\[\begin{aligned} \dfrac{3 x+12 y}{3} &=\dfrac{-24}{3} \\ x+4 y &=-8 \end{aligned} \nonumber \]

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 $\PageIndex{2}$

Given the graph of the line in Figure $\PageIndex{2}$, ﬁnd the equation Given the graph of the line below, ﬁnd the equation of the line in standard form.

fig 3.6.2.png — Figure $\PageIndex{2}$: Determine the equation of the line.

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 $\Delta y / \Delta x=-5 / 2$ (see Figure $\PageIndex{3}$). Substitute $−5/2$ form and $−3$ for $b$ in the slope-intercept form of the line.

fig 3.6.3.png — Figure $\PageIndex{3}$: The line has $y$-intercept $(0,−3)$ and slope $−5/2$.

\[\begin{aligned} y &= mx+b \quad \color{Red} \text { Slope-intercept form. } \\ y &= -\dfrac{5}{2} x-3 \quad \color{Red} \text { Substitute: }-5 / 2 \text { for } m,-3 \text { for } b \end{aligned} \nonumber \]

To put this result in standard form $Ax + By = C$, ﬁrst clear the fractions by multiplying both sides by the common denominator.

\[\begin{aligned} 2y &= 2\left[-\dfrac{5}{2} x-3\right] \quad \color{Red} \text { Multiply both sides by } 2 \\ 2y &= 2\left[-\dfrac{5}{2} x\right]-2[3] \quad \color{Red} \text { Distribute the } 2 \\ 2y &= -5x-6 \quad \color{Red} \text { Multiply. } \end{aligned} \nonumber \]

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.

\[\begin{aligned} 5x+2y &= -5x-6+5x \quad \color{Red} \text { Add } 5x \text { to both sides. } \\ 5x+2y &= -6 \quad \color{Red}\text { Simplify. } \end{aligned} \nonumber \]

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 $\PageIndex{2}$

Given the graph of the line below, ﬁnd the equation of the line in standard form.

$Exercise 3.6.2.png$

Answer: $3 x−4y = −2$

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 $\PageIndex{3}$

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 $\PageIndex{4}$).

fig 3.6.4.png — Figure $\PageIndex{4}$: The line through $(−3,−4)$ and $(1,2)$.

Use the points $(−3,−4)$ and $(1,2)$ to calculate the slope.

\[\begin{aligned} \text { Slope } &= \dfrac{\Delta y}{\Delta x} \quad \color{Red} \text { Slope formula. } \\ &= \dfrac{2-(-4)}{1-(-3)} \quad \color{Red} \text { Subtract coordinates of }(-3,-4) \\ &= \dfrac{6}{4} \quad \text { Simplify. } \\ &= \dfrac{3}{2} \quad \text { Reduce. } \end{aligned} \nonumber \]

Let’s substitute $\left(x_{0}, y_{0}\right)=(1,2)$ and $m =3 /2$ in the point-slope form of the line. (Note: Substituting $\left(x_{0}, y_{0}\right)=(-3,-4)$ and $m =3 /2$ would yield the same answer.)

\[\begin{aligned} y-y_{0} &= m\left(x-x_{0}\right) \quad \color{Red} \text { Point-slope form. } \\ y-2 &= \dfrac{3}{2}(x-1) \quad \color{Red} \text { Substitute: } 3 / 2 \text { for } m, 1 \text { for } x_{0} \end{aligned} \nonumber \]

The question requests that our ﬁnal answer be presented in standard form. First we clear the fractions.

\[\begin{aligned} y-2 &= \dfrac{3}{2} x-\dfrac{3}{2} \quad \color{Red} \text { Distribute the } 3 / 2 \\ 2[y-2] &= 2\left[\dfrac{3}{2} x-\dfrac{3}{2}\right] \quad \color{Red} \text { Multiply both sides by } 2 \\ 2y-2[2] &= 2\left[\dfrac{3}{2} x\right]-2\left[\dfrac{3}{2}\right] \quad \color{Red} \text { Distribute the } 2 \\ 2y-4 &= 3 x-3 \quad \color{Red} \text { Multiply. } \end{aligned} \nonumber \]

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.

\[\begin{aligned} 2y-4-3x &= 3x-3-3x \quad \color{Red} \text { Subtract } 3 x \text { from both sides. } \\ -3x+2y-4 &= -3 \quad \color{Red} \text { Simplify, changing the order on the left-hand side. } \end{aligned} \nonumber \]

To put this in the form $Ax + By = C$, we need to move the term $−4$ to the other side of the equation.

\[\begin{aligned} -3x+2y-4+4 &= -3+4 \quad \color{Red} \text { Add } 4 \text { to both sides. } \\ -3x+2y &= 1 \quad \color{Red} \text { Simplify. } \end{aligned} \nonumber \]

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 $−1$.

\[\begin{aligned} -1[-3x+2y] &= -1[1] \quad \color{Red} \text { Multiply both sides by }-1 \\ 3x-2y &= -1 \quad \color{Red} \text { Distribute the }-1 \end{aligned} \nonumber \]

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: \[y-2=\dfrac{6}{4}(x-1) \nonumber \]

Multiplying both sides by $4$ would give us the result \[4y−8=6x−6 \nonumber \]or equivalently: \[−6x +4y =2 \nonumber \]

This doesn’t look like the same answer, but if we divide both sides by $−2$, we do get the same result. \[3x−2y = −1 \nonumber \]

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 $\PageIndex{3}$

Find the standard form of the equation of the line that passes through the points $(−2,4)$ and $(3,−3)$.

Answer: $7x +5y =6$

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.

fig 3.6.5.png — Figure $\PageIndex{5}$: Each $x$-intercept has a $y$-coordinate equal to zero.

In Figure $\PageIndex{5}$, 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.

$x$ Intercepts: To ﬁnd the $x$-intercepts of the graph of an equation, substitute $y = 0$ into the equation and solve for $x$.

fig 3.6.6.png — Figure $\PageIndex{6}$: Each $y$-intercept has an $x$-coordinate equal to zero.

Similarly, the graph in Figure $\PageIndex{6}$ 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.

$y$ Intercepts: 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 $\PageIndex{4}$

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 $x$.

\[\begin{aligned} 2 x-3 y &=6 \\ 2 x-3(0) &=6 \\ 2 x &=6 \\ \dfrac{2 x}{2} &=\dfrac{6}{2} \\ x &=3 \end{aligned} \nonumber \]

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 $y$.

\[\begin{aligned} 2 x-3 y &=6 \\ 2(0)-3 y &=6 \\-3 y &=6 \\ \frac{-3 y}{-3} &=\frac{6}{-3} \\ y &=-2 \end{aligned} \nonumber \]

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 $\PageIndex{7}$).

fig 3.6.7.png — Figure $\PageIndex{7}$: The graph of $2x−3y = 6$ has intercepts $(3,0)$ and $(0,−2)$.

Exercise $\PageIndex{4}$

Find the $x$- and $y$-intercepts of the line having equation $3x +4y = −12$. Plot the intercepts and draw the line.

Answer

$x$-intercept: $(−4,0)$

$y$-intercept: $(0,−3)$

$Exercise 3.6.4.png$

Example $\PageIndex{5}$

Sketch the line $4x +3 y = 12$, then sketch the line through the point $(−2,−2)$ that is perpendicular to the line $4x +3 y = 12$. 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 $x$.

\[\begin{aligned} 4 x+3 y &=12 \\ 4 x+3(0) &=12 \\ 4 x &=12 \\ \dfrac{4 x}{4} &=\dfrac{12}{4} \\ x &=3 \end{aligned} \nonumber \]

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 $y$.

\[\begin{aligned} 4 x+3 y &=12 \\ 4(0)+3 y &=12 \\ 3 y &=12 \\ \dfrac{3 y}{3} &=\dfrac{12}{3} \\ y &=4 \end{aligned} \nonumber \]

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 $\PageIndex{8}$).

fig 3.6.8.png — Figure $\PageIndex{8}$: The graph of $4x+3y = 12$ has intercepts $(3,0)$ and $(0,4)$ and slope $−4/3$.

You could also solve for $y$ to put $3x +4 y = 12$ 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 $\PageIndex{9}$).

fig 3.6.9.png — Figure $\PageIndex{9}$: The slope of the perpendicular line is the negative reciprocal of $−4/3$, namely $3/4$.

Finally, use the point-slope form, $m =3 /4$, and $\left(x_{0}, y_{0}\right)=(-2,-2)$ to determine the equation of the perpendicular line.

\[\begin{aligned} y-y_{0} &= m\left(x-x_{0}\right) \quad \color{Red} \text { Point-slope form. } \\ y-(-2) &= \dfrac{3}{4}(x-(-2)) \quad \color{Red} \text { Substitute: } 3 / 4 \text { for } m,-2 \text { for } x_{0} \text { and }-2 \text { for } y_{0} \\ y+2 &= \dfrac{3}{4}(x+2) \quad \color{Red} \text { Simplify. } \end{aligned} \nonumber \]

Let’s place our answer in standard form. Clear the fractions.

\[\begin{aligned} y+2 &= \dfrac{3}{4} x+\dfrac{6}{4} \quad \color{Red} \text { Distribute } 3 / 4 \\ 4[y+2] &= 4\left[\dfrac{3}{4} x+\dfrac{6}{4}\right] \quad \color{Red} \text { Multiply both sides by } 4 \\ 4y+4[2] &= 4\left[\dfrac{3}{4} x\right]+4\left[\dfrac{6}{4}\right] \quad \color{Red} \text { Distribute the } 4 \\ 4y+8 &= 3x+6 \quad \color{Red} \text { Multiply. } \end{aligned} \nonumber \]

Rearrange the terms to put them in the order $Ax + By = C$.

\[\begin{aligned} 4y+8-3x &= 3x+6-3x \quad \color{Red} \text { Subtract } 3x \text { from both sides. } \\ -3x+4y+8 &= 6 \quad \color{Red} \text { Simplify. Rearrange on the left. } \\ -3x+4y+8-8 &= 6-8 \quad \color{Red} \text { Subtract } 8 \text { from both sides. } \\ -3x+4y &= -2 \quad \color{Red} \text { Simplify. } \\ -1(-3x+4y) &= -1(-2) \quad \color{Red} \text { Multiply both sides by }-1 \\ 3x-4y &= 2 \quad \color{Red} \text { Distribute the }-1 \end{aligned} \nonumber \]

Hence, the equation of the perpendicular line is $3x−4y = 2$.

Exercise $\PageIndex{5}$

Find the equation of the line that passes through the point $(3,2)$ and is perpendicular to the line $6x−5y = 15$.

Answer: $5x +6y = 27$

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, $3 x = 6$ can be simpliﬁed to $x = 2$ and $2 y = −12$ 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 $\PageIndex{6}$

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 $\PageIndex{10}$.

fig 3.6.10.png — Figure $\PageIndex{10}$: The graph of $x = 3$ is a vertical line.

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 $\PageIndex{11}$.

fig 3.6.11.png — Figure $\PageIndex{11}$: The graph of $y =−3$ is a horizontal line.

Things to note:

A couple of comments are in order regarding the lines in Figures $\PageIndex{10}$ and $\PageIndex{11}$.

The graph of $x = 3$ in Figure $\PageIndex{10}$, being a vertical line, has undeﬁned slope. Therefore, we cannot use either of the formulae $y = mx + b$ or $y−y_0 = m(x−x_0)$ 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:\[\begin{aligned} y &= mx+b \quad \color{Red} \text { Slope-intercept form. } \\ y &= 0x+(-3) \quad \color{Red} \text { Substitute: } 0 \text { for } m,-3 \text { for } b \\ y &= -3 \quad \color{Red} \text { Simplify. } \end{aligned} \nonumber \]

However, it is far easier to just look at the line in Figures $\PageIndex{11}$ and note that it is the collection of all points $(x,y)$ with $y = 3$.

Exercise $\PageIndex{6}$

Sketch the graphs of $x = −2$ and $y = 2$.

Answer: $Exercise 3.6.6.png$