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Understand Slope of a Line
Use geoboards to model slope
Use Geoboards to Model Slope
In this section, we will explore the concepts of slope.
Using rubber bands on a geoboard gives a concrete way to model lines on a coordinate grid. By stretching a rubber band between two pegs on a geoboard, we can discover how to find the slope of a line.
We’ll start by stretching a rubber band between two pegs to make a line as shown in .
Does it look like a line?
Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in . We carefully make a \(90^{\circ}\) angle around the third peg, so that one side is vertical and the other is horizontal.
To find the slope of the line, we measure the distance along the vertical and horizontal legs of the triangle. The vertical distance is called the rise and the horizontal distance is called the run, as shown in .
To help remember the terms, it may help to think of the images shown in .
\[\ m=\frac{\text{rise}}{\text{run}}\]\[m=\frac{2}{3}\]\[\text{The line has slope}\ \frac{2}{3}.\]Example
Try it.
What is the slope of the line on the geoboard shown?
Solution
Use the definition of slope.
\(m=\frac{\text{rise}}{\text{run}}\)
Start at the left peg and make a right triangle by stretching the rubber band up and to the right to reach the second peg.
Count the rise and the run as shown.
\(\begin{array}{llll}\text{The rise is}\ 3\ \text{units}. & & & m=\frac{3}{\text{run}} \\ \text{The run is}\ 4\ \text{units}. & & & m=\frac{3}{4} \\ & & & \text{The slope is}\ \frac{3}{4}.\end{array}\)
Example
Try it.
What is the slope of the line on the geoboard shown?
Solution
Use the definition of slope.
\(m=\frac{\text{rise}}{\text{run}}\)
Start at the left peg and make a right triangle by stretching the rubber band to the peg on the right. This time we need to stretch the rubber band down to make the vertical leg, so the rise is negative.
\(\begin{array}{llll}\text{The rise is}\ -1. & & & m=\frac{-1}{\text{run}} \\ \text{The run is}\ 3. & & & m=\frac{-1}{3} \\ & & & m=-\frac{1}{3} \\ & & & \text{The slope is}\ -\frac{1}{3}.\end{array}\)
Condensed — the full section is in OpenStax Prealgebra 2e.
Find the Slope of a Line from its Graph
Now we’ll look at some graphs on a coordinate grid to find their slopes. The method will be very similar to what we just modeled on our geoboards.
To find the slope, we must count out the rise and the run. But where do we start?
We locate any two points on the line. We try to choose points with coordinates that are integers to make our calculations easier. We then start with the point on the left and sketch a right triangle, so we can count the rise and run.
Example
Try it.
Find the slope of the line shown:
Solution
Locate two points on the graph, choosing points whose coordinates are integers. We will use \((0,-3)\) and \((5,1).\)
Starting with the point on the left, \((0,-3),\) sketch a right triangle, going from the first point to the second point, \((5,1).\)
| Count the rise on the vertical leg of the triangle. | The rise is 4 units. |
| Count the run on the horizontal leg. | The run is 5 units. |
| Use the slope formula. | \(m=\frac{\text{rise}}{\text{run}}\) |
| Substitute the values of the rise and run. | \(m=\frac{4}{5}\) |
| The slope of the line is \(\frac{4}{5}\). |
Notice that the slope is positive since the line slants upward from left to right.
The lines in the previous examples had \(y\)-intercepts with integer values, so it was convenient to use the y-intercept as one of the points we used to find the slope. In the next example, the \(y\)-intercept is a fraction. The calculations are easier if we use two points with integer coordinates.
Example
Try it.
Find the slope of the line shown:
Solution
| Locate two points on the graph whose coordinates are integers. | \((2,3)\) and \((7,6)\) |
| Which point is on the left? | \((2,3)\) |
| Starting at \((2,3)\), sketch a right angle to \((7,6)\) as shown below. |
| Count the rise. | The rise is 3. |
| Count the run. | The run is 5. |
| Use the slope formula. | \(m=\frac{\text{rise}}{\text{run}}\) |
| Substitute the values of the rise and run. | \(m=\frac{3}{5}\) |
| The slope of the line is \(\frac{3}{5}.\) |
Condensed — the full section is in OpenStax Prealgebra 2e.
Find the Slope of Horizontal and Vertical Lines
Do you remember what was special about horizontal and vertical lines? Their equations had just one variable.
- horizontal line \(y=b;\) all the \(y\)-coordinates are the same.
- vertical line \(x=a;\) all the \(x\)-coordinates are the same.
So how do we find the slope of the horizontal line \(y=4?\) One approach would be to graph the horizontal line, find two points on it, and count the rise and the run. Let’s see what happens in . We’ll use the two points \((0,\ 4)\) and \((3,\ 4)\) to count the rise and run.
| What is the rise? | The rise is 0. |
| What is the run? | The run is 3. |
| What is the slope? | \(m=\frac{\text{rise}}{\text{run}}\) |
| \(m=\frac{0}{3}\) | |
| \(m=0\) |
The slope of the horizontal line \(y=4\) is \(0.\)
All horizontal lines have slope \(0\). When the \(y\)-coordinates are the same, the rise is \(0\).
Now we’ll consider a vertical line, such as the line \(x=3\), shown in . We’ll use the two points \((3,\ 0)\) and \((3,\ 2)\) to count the rise and run.
| What is the rise? | The rise is 2. |
| What is the run? | The run is 0. |
| What is the slope? | \(m=\frac{\text{rise}}{\text{run}}\) |
| \(m=\frac{2}{0}\) |
But we can’t divide by \(0.\) Division by \(0\) is undefined. So we say that the slope of the vertical line \(x=3\) is undefined. The slope of all vertical lines is undefined, because the run is \(0.\)
Example
Try it.
Find the slope of each line:
- ⓐ \(\ x=8\\)
- ⓑ \(\ y=-5\)
Solution
ⓐ \(\ x=8\)
This is a vertical line, so its slope is undefined.
ⓑ \(\ y=-5\)
This is a horizontal line, so its slope is \(0.\)
Use the Slope Formula to find the Slope of a Line between Two Points
Sometimes we need to find the slope of a line between two points and we might not have a graph to count out the rise and the run. We could plot the points on grid paper, then count out the rise and the run, but there is a way to find the slope without graphing.
Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair \((x,y)\) gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol \((x,y)\) be used to represent two different points?
Mathematicians use subscripts to distinguish between the points. A subscript is a small number written to the right of, and a little lower than, a variable.
- \(({x}_{1},{y}_{1})\ \text{read}\ x\ \text{sub}\ 1,y\ \text{sub}\ 1\)
- \(({x}_{2},{y}_{2})\ \text{read}\ x\ \text{sub}\ 2,y\ \text{sub}\ 2\)
We will use \(({x}_{1},{y}_{1})\) to identify the first point and \(({x}_{2},{y}_{2})\) to identify the second point. If we had more than two points, we could use \(({x}_{3},{y}_{3}),({x}_{4},{y}_{4}),\) and so on.
To see how the rise and run relate to the coordinates of the two points, let’s take another look at the slope of the line between the points \((2,3)\) and \((7,6)\) in .
Since we have two points, we will use subscript notation.
\[\overset{{x}_{1},{y}_{1}}{(2,3)}\ \overset{{x}_{2},{y}_{2}}{(7,6)}\]On the graph, we counted the rise of \(3.\) The rise can also be found by subtracting the \(y\text{-coordinates}\) of the points.
\[\begin{array}{l}{y}_{2}-{y}_{1} \\ 6-3 \\ 3\end{array}\]\[\begin{array}{l}{x}_{2}-{x}_{1} \\ 7-2 \\ 5\end{array}\]| We know | \(m=\frac{\text{rise}}{\text{run}}\) |
| So | \(m=\frac{3}{5}\) |
| We rewrite the rise and run by putting in the coordinates. | \(m=\frac{6-3}{7-2}\) |
| But 6 is the \(y\)-coordinate of the second point, \({y}_{2}\) and 3 is the \(y\)-coordinate of the first point \({y}_{1}\). So we can rewrite the rise using subscript notation. | \(m=\frac{{y}_{2}-{y}_{1}}{7-2}\) |
| Also 7 is the \(x\)-coordinate of the second point, \({x}_{2}\) and 2 is the \(x\)-coordinate of the first point \({x}_{2}\). So we rewrite the run using subscript notation. | \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) |
Condensed — the full section is in OpenStax Prealgebra 2e.
Graph a Line Given a Point and the Slope
In this chapter, we graphed lines by plotting points, by using intercepts, and by recognizing horizontal and vertical lines.
Another method we can use to graph lines is the point-slope method. Sometimes, we will be given one point and the slope of the line, instead of its equation. When this happens, we use the definition of slope to draw the graph of the line.
Example
Try it.
Graph the line passing through the point \((1,-1)\) whose slope is \(m=\frac{3}{4}.\)
Solution
Plot the given point, \((1,-1).\)
Use the slope formula \(m=\frac{\text{rise}}{\text{run}}\) to identify the rise and the run.
\[\begin{array}{l} \\ \\ m=\frac{3}{4} \\ \frac{\text{rise}}{\text{run}}=\frac{3}{4} \\ \\ \\ \text{rise}=3 \\ \text{run}=4\end{array}\]Starting at the point we plotted, count out the rise and run to mark the second point. We count \(3\) units up and \(4\) units right.
Then we connect the points with a line and draw arrows at the ends to show it continues.
We can check our line by starting at any point and counting up \(3\) and to the right \(4.\) We should get to another point on the line.
Example
Try it.
Graph the line with \(y\)-intercept \((0,2)\) and slope \(m=-\frac{2}{3}.\)
Solution
Plot the given point, the \(y\)-intercept \((0,2).\)
Use the slope formula \(m=\frac{\text{rise}}{\text{run}}\) to identify the rise and the run.
\[\begin{array}{l} \\ \\ m=-\frac{2}{3} \\ \frac{\text{rise}}{\text{run}}=\frac{-2}{3} \\ \\ \\ \text{rise}=-2 \\ \text{run}=3\end{array}\]Starting at \((0,2),\) count the rise and the run and mark the second point.
Connect the points with a line.
Example
Try it.
Graph the line passing through the point \((-1,-3)\) whose slope is \(m=4.\)
Solution
Plot the given point.
| Identify the rise and the run. | \(m=4\) |
| Write 4 as a fraction. | \(\frac{\text{rise}}{\text{run}}=\frac{4}{1}\) |
| \(\text{rise}=4\ \text{run}=1\) |
Count the rise and run.
Mark the second point. Connect the two points with a line.
Condensed — the full section is in OpenStax Prealgebra 2e.
Solve Slope Applications
At the beginning of this section, we said there are many applications of slope in the real world. Let’s look at a few now.
Example
Try it.
The pitch of a building’s roof is the slope of the roof. Knowing the pitch is important in climates where there is heavy snowfall. If the roof is too flat, the weight of the snow may cause it to collapse. What is the slope of the roof shown?
Solution
| Use the slope formula. | \(m=\frac{\text{rise}}{\text{run}}\) |
| Substitute the values for rise and run. | \(m=\frac{\text{9 ft}}{\text{18 ft}}\) |
| Simplify. | \(m=\frac{1}{2}\) |
| The slope of the roof is \(\frac{1}{2}\). |
Have you ever thought about the sewage pipes going from your house to the street? Their slope is an important factor in how they take waste away from your house.
Example
Try it.
Sewage pipes must slope down \(\frac{1}{4}\) inch per foot in order to drain properly. What is the required slope?
Solution
| Use the slope formula. | \(m=\frac{\text{rise}}{\text{run}}\) |
| \(m=\frac{-\frac{1}{4}\ \text{in}\text{.}}{1\ \text{ft}}\) | |
| \(m=\frac{-\frac{1}{4}\ \text{in}\text{.}}{1\ \text{ft}}\) | |
| Convert 1 foot to 12 inches. | \(m=\frac{-\frac{1}{4}\ \text{in}\text{.}}{12\ \text{in.}}\) |
| Simplify. | \(m=-\frac{1}{48}\) |
| The slope of the pipe is \(-\frac{1}{48}.\) |
Key Concepts
- Find the slope from a graph
- Locate two points on the line whose coordinates are integers.
- Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point.
- Count the rise and the run on the legs of the triangle.
- Take the ratio of rise to run to find the slope, \(m=\frac{\text{rise}}{\text{run}}\)
- Slope of a Horizontal Line
- The slope of a horizontal line, \(y=b\), is 0.
- Slope of a Vertical Line
- The slope of a vertical line, \(x=a\), is undefined.
- Slope Formula
- The slope of the line between two points \(({x}_{1},{y}_{1})\) and \(({x}_{2},{y}_{2})\) is \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)
- Graph a line given a point and a slope.
- Plot the given point.
- Use the slope formula to identify the rise and the run.
- Starting at the given point, count out the rise and run to mark the second point.
- Connect the points with a line.
Chapter Practice Test
In the following exercises, find three solutions to each equation and then graph each line.
In the following exercises, find the slope of each line.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Simplify: \(\frac{1-4}{8-2}.\)
If you missed this problem, review .جواب رو نشون بده
\(-\frac{1}{2}\)
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Divide: \(\frac{0}{4},\frac{4}{0}.\)
If you missed this problem, review .جواب رو نشون بده
\(\text{0, undefined}\)
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Simplify: \(\frac{15}{-3},\frac{-15}{3},\frac{-15}{-3}.\)
If you missed this problem, review .جواب رو نشون بده
\(-5,-5,5\)
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What is the slope of the line on the geoboard shown?
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Use the definition of slope.
\(m=\frac{\text{rise}}{\text{run}}\)
Start at the left peg and make a right triangle by stretching the rubber band up and to the right to reach the second peg.
Count the rise and the run as shown.
\(\begin{array}{llll}\text{The rise is}\ 3\ \text{units}. & & & m=\frac{3}{\text{run}} \\ \text{The run is}\ 4\ \text{units}. & & & m=\frac{3}{4} \\ & & & \text{The slope is}\ \frac{3}{4}.\end{array}\)
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What is the slope of the line on the geoboard shown?
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\(\frac{4}{3}\)
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What is the slope of the line on the geoboard shown?
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\(\frac{1}{4}\)
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What is the slope of the line on the geoboard shown?
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Use the definition of slope.
\(m=\frac{\text{rise}}{\text{run}}\)
Start at the left peg and make a right triangle by stretching the rubber band to the peg on the right. This time we need to stretch the rubber band down to make the vertical leg, so the rise is negative.
\(\begin{array}{llll}\text{The rise is}\ -1. & & & m=\frac{-1}{\text{run}} \\ \text{The run is}\ 3. & & & m=\frac{-1}{3} \\ & & & m=-\frac{1}{3} \\ & & & \text{The slope is}\ -\frac{1}{3}.\end{array}\)
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What is the slope of the line on the geoboard?
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\(-\frac{2}{3}\)
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What is the slope of the line on the geoboard?
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\(-\frac{4}{3}\)
-
Use a geoboard to model a line with slope \(\frac{1}{2}.\)
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To model a line with a specific slope on a geoboard, we need to know the rise and the run.
Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Replace \(m\) with \(\frac{1}{2}\). \(\frac{1}{2}=\frac{\text{rise}}{\text{run}}\) So, the rise is \(1\) unit and the run is \(2\) units.
Start at a peg in the lower left of the geoboard. Stretch the rubber band up \(1\) unit, and then right \(2\) units.
The hypotenuse of the right triangle formed by the rubber band represents a line with a slope of \(\frac{1}{2}.\)
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Use a geoboard to model a line with the given slope: \(m=\frac{1}{3}.\)
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Use a geoboard to model a line with the given slope: \(m=\frac{3}{2}.\)
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Use a geoboard to model a line with slope \(\frac{-1}{4},\)
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Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Replace \(m\) with \(-\frac{1}{4}\). \(-\frac{1}{4}=\frac{\text{rise}}{\text{run}}\) So, the rise is \(-1\) and the run is \(4.\)
Since the rise is negative, we choose a starting peg on the upper left that will give us room to count down. We stretch the rubber band down \(1\) unit, then to the right \(4\) units.
The hypotenuse of the right triangle formed by the rubber band represents a line whose slope is \(-\frac{1}{4}.\)
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Use a geoboard to model a line with the given slope: \(m=\frac{-2}{1}.\)
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Use a geoboard to model a line with the given slope: \(m=\frac{-1}{3}.\)
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-
Find the slope of the line shown:
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Locate two points on the graph, choosing points whose coordinates are integers. We will use \((0,-3)\) and \((5,1).\)
Starting with the point on the left, \((0,-3),\) sketch a right triangle, going from the first point to the second point, \((5,1).\)
Count the rise on the vertical leg of the triangle. The rise is 4 units. Count the run on the horizontal leg. The run is 5 units. Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Substitute the values of the rise and run. \(m=\frac{4}{5}\) The slope of the line is \(\frac{4}{5}\). Notice that the slope is positive since the line slants upward from left to right.
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Find the slope of the line:
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\(\frac{2}{5}\)
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Find the slope of the line:
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\(\frac{3}{4}\)
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Find the slope of the line shown:
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Locate two points on the graph. Look for points with coordinates that are integers. We can choose any points, but we will use (0, 5) and (3, 3). Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point.
Count the rise – it is negative. The rise is −2. Count the run. The run is 3. Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Substitute the values of the rise and run. \(m=\frac{-2}{3}\) Simplify. \(m=-\frac{2}{3}\) The slope of the line is \(-\frac{2}{3}.\) Notice that the slope is negative since the line slants downward from left to right.
What if we had chosen different points? Let’s find the slope of the line again, this time using different points. We will use the points \((-3,7)\) and \((6,1).\)
Starting at \((-3,7),\) sketch a right triangle to \((6,1).\)
Count the rise. The rise is −6. Count the run. The run is 9. Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Substitute the values of the rise and run. \(m=\frac{-6}{9}\) Simplify the fraction. \(m=-\frac{2}{3}\) The slope of the line is \(-\frac{2}{3}.\) It does not matter which points you use—the slope of the line is always the same. The slope of a line is constant!
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Find the slope of the line:
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\(-\frac{4}{3}\)
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Find the slope of the line:
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\(-\frac{3}{5}\)
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Find the slope of the line shown:
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Locate two points on the graph whose coordinates are integers. \((2,3)\) and \((7,6)\) Which point is on the left? \((2,3)\) Starting at \((2,3)\), sketch a right angle to \((7,6)\) as shown below.
Count the rise. The rise is 3. Count the run. The run is 5. Use the slope formula. \(m=\frac{\text{rise}}{\text{run}}\) Substitute the values of the rise and run. \(m=\frac{3}{5}\) The slope of the line is \(\frac{3}{5}.\) -
Find the slope of the line:
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\(\frac{5}{4}\)
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Find the slope of the line:
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\(\frac{3}{2}\)
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Find the slope of each line:
- ⓐ \(\ x=8\\)
- ⓑ \(\ y=-5\)
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ⓐ \(\ x=8\)
This is a vertical line, so its slope is undefined.
ⓑ \(\ y=-5\)
This is a horizontal line, so its slope is \(0.\)
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Find the slope of the line: \(x=-4.\)
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undefined
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Find the slope of the line: \(y=7.\)
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0
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Find the slope of the line between the points \((1,2)\) and \((4,5).\)
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We’ll call \((1,2)\) point #1 and \((4,5)\)point #2. \(\overset{{x}_{1},{y}_{1}}{(1,2)}\ \text{and}\ \overset{{x}_{2},{y}_{2}}{(4,5)}\) Use the slope formula. \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) Substitute the values in the slope formula: \(y\\) of the second point minus \(\ y\\) of the first point \(m=\frac{5-2}{{x}_{2}-{x}_{1}}\) \(x\\) of the second point minus \(\ x\\) of the first point \(m=\frac{5-2}{4-1}\) Simplify the numerator and the denominator. \(m=\frac{3}{3}\) \(m=1\) Let’s confirm this by counting out the slope on the graph.
The rise is \(3\) and the run is \(3,\) so
\(\begin{array}{l} \\ m=\frac{\text{rise}}{\text{run}} \\ m=\frac{3}{3} \\ m=1\end{array}\)
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Find the slope of the line through the given points: \((8,5)\) and \((6,3).\)
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1
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Find the slope of the line through the given points: \((1,5)\) and \((5,9)\text{.}\)
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1
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Find the slope of the line through the points \((-2,-3)\) and \((-7,4).\)
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We’ll call \((-2,-3)\) point #1 and \((-7,4)\) point #2. \(\overset{{x}_{1},{y}_{1}}{(-2,-3)}\ \text{and}\ \overset{{x}_{2},{y}_{2}}{(-7,4)}\) Use the slope formula. \(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\) Substitute the values \(y\\) of the second point minus \(\ y\\) of the first point \(m=\frac{4-(-3)}{{x}_{2}-{x}_{1}}\) \(x\\) of the second point minus \(\ x\\) of the first point \(m=\frac{4-(-3)}{-7-(-2)}\) Simplify. \(m=\frac{7}{-5}\) \(m=-\frac{7}{5}\) Let’s confirm this on the graph shown.
\(\begin{array}{l} \\ \\ \\ m=\frac{\text{rise}}{\text{run}} \\ m=\frac{-7}{5} \\ m=-\frac{7}{5}\end{array}\)
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Find the slope of the line through the pair of points: \((-3,4)\) and \((2,-1).\)
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−1
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Find the slope of the line through the pair of points: \((-2,6)\) and \((-3,-4).\)
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10
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Graph the line passing through the point \((1,-1)\) whose slope is \(m=\frac{3}{4}.\)
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Plot the given point, \((1,-1).\)
Use the slope formula \(m=\frac{\text{rise}}{\text{run}}\) to identify the rise and the run.
\[\begin{array}{l} \\ \\ m=\frac{3}{4} \\ \frac{\text{rise}}{\text{run}}=\frac{3}{4} \\ \\ \\ \text{rise}=3 \\ \text{run}=4\end{array}\]Starting at the point we plotted, count out the rise and run to mark the second point. We count \(3\) units up and \(4\) units right.
Then we connect the points with a line and draw arrows at the ends to show it continues.
We can check our line by starting at any point and counting up \(3\) and to the right \(4.\) We should get to another point on the line.
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Graph the line passing through the point with the given slope:
\((2,-2),m=\frac{4}{3}\)
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Graph the line passing through the point with the given slope:
\((-2,3),m=\frac{1}{4}\)
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Graph the line with \(y\)-intercept \((0,2)\) and slope \(m=-\frac{2}{3}.\)
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Plot the given point, the \(y\)-intercept \((0,2).\)
Use the slope formula \(m=\frac{\text{rise}}{\text{run}}\) to identify the rise and the run.
\[\begin{array}{l} \\ \\ m=-\frac{2}{3} \\ \frac{\text{rise}}{\text{run}}=\frac{-2}{3} \\ \\ \\ \text{rise}=-2 \\ \text{run}=3\end{array}\]Starting at \((0,2),\) count the rise and the run and mark the second point.
Connect the points with a line.
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Graph the line with the given intercept and slope:
\(y\)-intercept \(4,m=-\frac{5}{2}\)
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Graph the line with the given intercept and slope:
\(x\)-intercept \(-3,m=-\frac{3}{4}\)
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Graph the line passing through the point \((-1,-3)\) whose slope is \(m=4.\)
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Plot the given point.
Identify the rise and the run. \(m=4\) Write 4 as a fraction. \(\frac{\text{rise}}{\text{run}}=\frac{4}{1}\) \(\text{rise}=4\ \text{run}=1\) Count the rise and run.
Mark the second point. Connect the two points with a line.
Symbols used here
1/360 of a full turn. 180° = π radians.
Both signs at once: x = 3 ± 2 means 5 and 1.
The two sides are different.
Inequalities that allow equality; < and > exclude it.
Equal to the precision shown, not exactly.
The non-negative number whose square (n-th power) is x.
What is left after dividing a by n.
Per hundred: 15% = 15/100.
a for every b; a divided by b.
How to: Understand Slope of a Line
- Use geoboards to model slope
- Find the slope of a line from its graph
- Find the slope of horizontal and vertical lines
- Use the slope formula to find the slope of a line between two points
- Graph a line given a point and the slope Solve slope applications
- Locate two points on the line whose coordinates are integers.
- Starting with the point on the left, sketch a right triangle, with the hypotenuse going from the first point to the second point.
- Count the rise and the run on the legs of the triangle.
Questions people ask
Why does the order of operations matter?
Because 2 + 3 × 4 would otherwise be two different numbers. The convention (brackets, exponents, multiplication and division, addition and subtraction) exists so every reader gets the same value from the same expression.
How do I check an arithmetic answer?
Estimate first (round every number and compute roughly), then compare. If the estimate and the exact answer disagree by more than a little, one of them is wrong. The solver shows every operation, so you can find which line went astray.
Why are fractions harder than decimals?
They are not harder, they are more exact: 1/3 is a precise number, 0.333 is an approximation. Fractions need a common denominator to add, which is the one extra step people trip on.
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Parts of this page are adapted from OpenStax Prealgebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.