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Section 6.2 Intercepts and the Standard Form of the Equation of a Line

The Standard Form of the Equation of a Line.

The standard form of the equation of a line is \(ax+by=c\) where \(a\text{,}\) \(b\text{,}\) and \(c\) represent real numbers. While either \(a\) or \(b\) can be zero, they cannot both be zero in the same equation of a line.

To determine points on the line, we replace one of the variables in the equation with a number and solve the resultant equation for the other variable. For example, consider the line with equation \(3x-4y=24\text{.}\) If we wanted to determine the point on the line that has an \(x\)-coordinate of \(4\) we would go through the following process.

\begin{align*} 3(4)-4y\amp=24\\ 12-4y\amp=24\\ 12-4y\subtractright{12}\amp=24\subtractright{12}\\ -4y\amp=12\\ \divideunder{-4y}{-4}\amp=\divideunder{12}{-4}\\ y\amp=-3 \end{align*}

We now know that the point is \((4,-3)\text{.}\)

Similarly, if we wanted to know the point on the line that has a \(y\)-coordinate of \(15\text{,}\) we would do the following.

\begin{align*} 3x-4(15)\amp=24\\ 3x-60\amp=24\\ 3x-60\addright{60}\amp=24\addright{60}\\ 3x\amp=84\\ \divideunder{3x}{3}\amp=\divideunder{84}{3}\\ x\amp=28 \end{align*}

We can conclude that the point is \((28,15)\text{.}\)

\(x\)-intercepts and \(y\)-intercepts.

The point on a line with an \(x\)-coordinate of zero (if such a point exists) is called the \(y\)-intercept of the line. Similarly, the point with a \(y\)-coordinate of zero is called the \(x\)-intercept of the line.

For example, the \(y\)-intercept of the line shown in Figure 6.2.1 is \((0,-5)\text{;}\) note that this is the point where the line intersects the \(y\)-axis. The \(x\)-intercept of the same line is \((-2,0)\text{,}\) the point where the line intersects the \(x\)-axis.

The graph of the line that passes through the points \((-2,0)\) and \((0,-5)\text{.}\)
Figure 6.2.1. The \(x\) and \(y\) intercepts
Example 6.2.2.

Determine the \(x\)-intercept of the line with equation \(-11x+5y=-110\text{.}\)

Solution

Since the \(x\)-intercept is a point on the \(x\)-axis, it must have a \(y\)-coordinate of zero. Replacing \(y\) with zero and solving for \(x\) results in \(x=10\text{.}\) So the \(x\)-intercept of the line is the point \((10,0)\text{.}\)

Example 6.2.3.

Determine the \(y\)-intercept of the line with equation \(-11x+5y=-110\text{.}\)

Solution

Since the \(y\)-intercept is a point on the \(y\)-axis, it must have an \(x\)-coordinate of zero. Replacing \(x\) with zero and solving for \(y\) results in \(y=-21\text{.}\) So the \(y\)-intercept of the line is the point \((0,-21)\text{.}\)

Figure 6.2.4. Determine the \(x\)-intercept and \(y\)-intercept of the line with the given equation.
Example 6.2.5.

Graph the line with equation \(2x-3y=-6\text{.}\)

Solution

When graphing a line by hand it's useful to have at least three points with which to align your ruler. Let's consider the line with equation \(2x-3y=-6\text{.}\) The \(x\)-intercept and \(y\)-intercept of this line are, respectively, \((3,0)\) and \((0,-2)\text{.}\) In the next section we discuss the idea of the slope of the line, but we can use the basic idea right now to determine a couple of more points on the line.

The line \(2x-3y=-6\) in shown in Figure 6.2.6. Note that one way to describe the movement from the \(y\)-intercept to the \(x\)-intercept is "up \(2\text{,}\) right \(3\text{.}\)" If we continue that pattern, starting at the \(x\)-intercept, we land at the point \((6,2)\text{,}\) which is also on the line.

A graph of the line that passes through the points \((-3,-4)\) and \((6,2)\text{.}\)  There is a stair-like pattern of arrows pointing up and right.  The first "step" starts at \((-2,0)\text{,}\) points up to  \((0,0)\text{,}\) and then points right to \((3,0)\text{.}\)  The second "step" starts at \((3,0)\text{,}\) points up to \((3,2)\text{,}\) and then points right to \((6,2)\text{.}\)  There is another stair-like pattern that points down and left.  The first "step" starts at \((3,0)\text{,}\) points down to \((3,-2)\text{,}\) and then points left to \((0,-2)\text{.}\)  The second step starts at \((0,-2)\text{,}\) points down to \((0,-4)\text{,}\) and then points left to \((-3,-4)\text{.}\)
Figure 6.2.6. The step-like pattern between points on a line

Similarly, the movement from the \(x\)-intercept to the \(y\)-intercept can be described as "down \(2\text{,}\) left \(3\text{,}\)" and continuing that pattern takes us to the point \((-3,-4)\) which is also a point on the line.

Every line has its on step-like pattern that can be executed either left-to-right or right-to-left. This phenomenon is the basis for the concept of "slope of a line" and is explored in-depth in the next section.

Figure 6.2.7. Drag the points so that the line matches the given equation. A message appears when you are correct.

Exercises Exercises

Determine the indicated ordered pairs (points) on the lines with the given equations.

1.

Complete the entries in Table 6.2.8 for the line with equation \(3x-y=6\text{.}\) Also, state the \(x\) and \(y\) intercepts of the line.

\(x\) \(y\)
\(-1\)
\(2\)
\(5\)
\(8\)
\(11\)
Table 6.2.8. \(3x-y=6\)
Solution

The \(x\)-intercept is \((2,0)\) and the \(y\)-intercept is \((0,-6)\text{.}\)

\(x\) \(y\)
\(-1\) \(-9\)
\(2\) \(0\)
\(5\) \(9\)
\(8\) \(18\)
\(11\) \(27\)
Table 6.2.9. \(3x-y=6\)
2.

Complete the entries in Table 6.2.10 for the line with equation \(7x+4y=-8\text{.}\) Also, state the \(x\) and \(y\) intercepts of the line.

\(x\) \(y\)
\(2\)
\(5\)
\(-2\)
\(\frac{4}{7}\)
\(-\frac{5}{4}\)
Table 6.2.10. \(7x+4y=-8\)
Solution

The \(x\)-intercept is \(\left(-\frac{8}{7},0\right)\) and the \(y\)-intercept is \((0,-2)\text{.}\)

\(x\) \(y\)
\(2\) \(-\frac{11}{2}\)
\(-4\) \(5\)
\(-2\) \(\frac{3}{2}\)
\(\frac{4}{7}\) \(-3\)
\(-\frac{3}{7}\) \(-\frac{5}{4}\)
Table 6.2.11. \(7x+4y=-8\)