Section3.3Slope and the slope-intercept form of linear equations
Click here to open the Desmos graph full screen.
Subsection3.3.1Activities and Problems
Recall that one equation for slope is \(m=\frac{\text{rise}}{\text{run}}\text{.}\)
Use the slider labeled \(m\) to set the slope to \(3\text{.}\) Suppose that the run along the slope triangle was \(2\text{.}\) What would the rise then be? Use the slider labeled \(n\) to adjust the run to \(2\text{.}\) If necessary, adjust the value of \(x_1\) so that you can see the slope triangle. Is the rise what you expected it to be?
SolutionThe rise would be \(6\text{.}\)
Two of the three elements of the slope equation \(m=\frac{\text{rise}}{\text{run}}\) are stated in each row of Table 3.3.1. Find the missing element for each row and check your result with the Desmos graph.
| \(slope\) | rise | run |
| \(\) | \(25\) | \(-5\) |
| \(4\) | \(\) | \(-2\) |
| \(8\) | \(4\) | \(\) |
| \(-2\) | \(-2\) | \(\) |
| \(-6\) | \(\) | \(\frac{1}{3}\) |
| \(slope\) | rise | run |
| \(-5\) | \(25\) | \(-5\) |
| \(4\) | \(-8\) | \(-2\) |
| \(8\) | \(4\) | \(\frac{1}{2}\) |
| \(-2\) | \(-2\) | \(1\) |
| \(-6\) | \(-2\) | \(\frac{1}{3}\) |
Adjust the slider labeled \(b\) so that the line crosses the \(y\)-axis somewhere close to the origin. Use the slider labeled \(m\) to manually change the slope. What is true about all lines with positive slope that is distinct from all lines with negative slope? What is true about a line when its slope is zero? Change the value of \(b\) and confirm your conjecture about positive, negative, and zero slopes.
SolutionAll lines with positive slope increase (go upward) from left-to-right whereas all lines with negative slope decrease (go downward) from left-to-right. All lines with a slope of zero are horizontal. Although you can't verify it with the Desmo graph, vertical lines have no slope (sometimes stated as an undefined slope).
For each row in Table 3.3.3, use the formula \(\frac{y_2-y_1}{x_2-x_1}\) to determine the slope of the line that passes through the two given points.
| \((x_1,y_1)\) | \((x_2,y_2)\) |
| \((5,11)\) | \((3,7)\) |
| \((3,7)\) | \((5,11)\) |
| \((-3,2)\) | \((1,-5)\) |
| \((1,-5)\) | \((-3,2)\) |
| \((\frac{3}{2},\frac{1}{3})\) | \((4,-\frac{8}{3})\) |
| \((4,-\frac{8}{3})\) | \((\frac{3}{2},\frac{1}{3})\) |
| \((x_1,y_1)\) | \((x_2,y_2)\) | slope |
| \((5,11)\) | \((3,7)\) | \(-2\) |
| \((3,7)\) | \((5,11)\) | \(-2\) |
| \((-3,2)\) | \((1,-5)\) | \(-\frac{7}{4}\) |
| \((1,-5)\) | \((-3,2)\) | \(-\frac{7}{4}\) |
| \((\frac{3}{2},\frac{1}{3})\) | \((4,-\frac{8}{3})\) | \(-\frac{6}{5}\) |
| \((4,-\frac{8}{3})\) | \((\frac{3}{2},\frac{1}{3})\) | \(-\frac{6}{5}\) |
Determine the slope of the line shown in Figure 3.3.5.
SolutionThe slope is \(\frac{3}{2}\text{.}\)
Determine the slope of the line shown in Figure 3.3.6.
SolutionThe slope is \(-\frac{5}{4}\text{.}\)
Determine \(y\)-intercept of the line shown in Figure 3.3.5.
SolutionThe \(y\)-intercept is \((0,2)\text{.}\)
Determine \(y\)-intercept of the line shown in Figure 3.3.6.
SolutionThe \(y\)-intercept is \((0,-3)\text{.}\)
Recall that the slope-intercept form of the equation of a line is \(y=mx+b\) where \(m\) is the slope of the line and \(b\) is the \(y\)-coordinate of the \(y\)-intercept of the line. Use this information to determine the slope-intercept form of the equation of the line shown in Figure 3.3.5.
SolutionThe equation is \(y=\frac{3}{2}x+2\text{.}\)
Determine the slope-intercept form of the equation of the line shown in Figure 3.3.6.
SolutionThe equation is \(y=-\frac{5}{4}x-3\text{.}\)
Determine the slope-intercept form of the equation of the line that passes through the points \((2,-2)\) and \((0,3)\text{.}\) Use the Desmos graph to confirm your equation.
SolutionWe begin by using the formula \(m=\frac{y_2-y_1}{x_2-x_1}\text{.}\)
\(m=\frac{3-(-2)}{0-2}=-\frac{5}{2}\)
We know that the \(y\)-intercept is \((0,3)\text{,}\) so the slope-intercept equation for the line is \(y=-\frac{5}{2}x+3\text{.}\)
Suppose that we were asked to determine the slope-intercept equation of the line that passes through the points \((3,-2)\) and \((-5,2)\text{.}\) What approach might we take?
We could begin by using the formula \(m=\frac{y_2-y_1}{x_2-x_1}\) to determine that the slope of the line is \(-\frac{1}{2}\text{.}\)
We don't directly know the \(y\)-intercept of the line, but from the slope we know that the equation of the line is \(y=-\frac{1}{2}x+b\) where \(b\) is the \(y\)-coordinate of the \(y\)-intercept. So we could substitute the coordinates of the ordered pair \((3,-2)\) into that equation which would give us \(-2=(-\frac{1}{2})(3)+b\text{,}\) and solving for \(b\) we would get \(-\frac{1}{2}\text{,}\) so the slope-intercept equation for the line is \(y=-\frac{1}{2}x-\frac{1}{2}\text{.}\) Note that we could also have used the ordered pair \((-5,2)\) to determine the value of \(b\text{.}\)
Now it's your turn. Determine the slope-intercept equation for the line that passes through the points shown in Table 3.3.7.
| \(x\) | \(y\) |
| \(2\) | \(-7\) |
| \(-3\) | \(8\) |
We begin by using the formula \(m=\frac{y_2-y_1}{x_2-x_1}\) to determine that the slope of the line is \(-3\text{.}\)
We don't directly know the \(y\)-intercept of the line, but from the slope we know that the equation of the line is \(y=-3x+b\) where \(b\) is the \(y\)-coordinate of the \(y\)-intercept. Substituting the coordinates of the ordered pair \((2,-7)\) into that equation gives us \(-7=(-3)(2)+b\text{,}\) and solving for \(b\) we get \(-1\text{,}\) so the slope-intercept equation for the line is \(y=-3x-1\text{.}\) Note that we could also have used the ordered pair \((-3,8)\) to determine the value of \(b\text{.}\)
Determine the slope-intercept equation for the line that passes through the points shown in Table 3.3.8.
| \(x\) | \(y\) |
| \(-3\) | \(5\) |
| \(5\) | \(1\) |
We begin by using the formula \(m=\frac{y_2-y_1}{x_2-x_1}\) to determine that the slope of the line is \(-\frac{1}{2}\text{.}\)
We don't directly know the \(y\)-intercept of the line, but from the slope we know that the equation of the line is \(y=-\frac{1}{2}x+b\) where \(b\) is the \(y\)-coordinate of the \(y\)-intercept. Substituting the coordinates of the ordered pair \((5,1)\) into that equation gives us \(1=-\frac{1}{2}(5)+b\text{,}\) and solving for \(b\) we get \(b=\frac{7}{2}\text{,}\) so the slope-intercept equation for the line is \(y=-\frac{1}{2}x+\frac{7}{2}\text{.}\) Note that we could also have used the ordered pair \((-3,5)\) to determine the value of \(b\text{.}\)
Determine the slope-intercept equation for the line that passes through the points shown in Table 3.3.9.
| \(x\) | \(y\) |
| \(\frac{1}{2}\) | \(5\) |
| \(-\frac{3}{2}\) | \(-3\) |
The equation of the line is \(y=4x+3\text{.}\)
Explain why the three points shown in Table 3.3.10 cannot possibly all lie on the same line.
| \(x\) | \(y\) |
| \(2\) | \(7\) |
| \(5\) | \(12\) |
| \(8\) | \(19\) |
The line that passes through the first two points has a slope of \(\frac{5}{3}\) whereas the line that passes through the first and third point has a slope of \(2\text{.}\) If the points all lied on the same line, we would get the same slope no matter which two points we used in the slope formula.