{"id":10698,"date":"2017-06-05T15:04:47","date_gmt":"2017-06-05T15:04:47","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/prealgebra\/?post_type=chapter&#038;p=10698"},"modified":"2020-10-22T09:16:18","modified_gmt":"2020-10-22T09:16:18","slug":"finding-slope-given-two-points-on-a-line","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/finding-slope-given-two-points-on-a-line\/","title":{"raw":"9.2.b - Using the Slope Formula to Find the Slope between Two Points","rendered":"9.2.b &#8211; Using the Slope Formula to Find the Slope between Two Points"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the slope formula to find the slope of a line between two points<\/li>\r\n \t<li>Find the slope of horizontal and vertical lines<\/li>\r\n<\/ul>\r\n<\/div>\r\nSometimes 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.\r\n\r\nBefore we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair [latex]\\left(x,y\\right)[\/latex] gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol [latex]\\left(x,y\\right)[\/latex] be used to represent two different points?\r\n\r\nMathematicians 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.\r\n\r\nWe will use [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] to identify the first point and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] to identify the second point. (If we had more than two points, we could use [latex]\\left({x}_{3},{y}_{3}\\right),\\left({x}_{4},{y}_{4}\\right)[\/latex], and so on.)\r\n<h2>The Slope Formula<\/h2>\r\nYou\u2019ve seen that you can find the slope of a line on a graph by measuring the rise and the run. You can also find the slope of a straight line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an\u00a0[latex]x[\/latex]-value and a [latex]y[\/latex]-value, written as an ordered pair [latex](x, y)[\/latex]. The [latex]x[\/latex] value tells you where a point is horizontally. The [latex]y[\/latex] value tells you where the point is vertically.\r\n\r\nConsider two points on a line\u2014Point 1 and Point 2. Point 1 has coordinates [latex]\\left(x_{1},y_{1}\\right)[\/latex]\u00a0and Point 2 has coordinates [latex]\\left(x_{2},y_{2}\\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064308\/image031.jpg\" alt=\"A line with its rise and run. The first point on the line is labeled Point 1, or (x1, y1). The second point on the line is labeled Point 2, or (x2,y2). The rise is (y2 minus y1). The run is (x2 minus X1).\" width=\"416\" height=\"401\" \/>\r\n\r\nThe rise is the vertical distance between the two points, which is the difference between their \u00a0[latex]y[\/latex]-coordinates. That makes the rise [latex]\\left(y_{2}-y_{1}\\right)[\/latex]. The run between these two points is the difference in the [latex]x[\/latex]-coordinates, or [latex]\\left(x_{2}-x_{1}\\right)[\/latex].\r\n\r\nSo, [latex] \\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex] or [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex].\r\n\r\nTo see how the rise and run relate to the coordinates of the two points, let\u2019s take another look at the slope of the line between the points [latex]\\left(2,3\\right)[\/latex] and [latex]\\left(7,6\\right)[\/latex] below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224620\/CNX_BMath_Figure_11_04_030.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6. It os labeled x sub 2 minus x sub 1, 7 minus 2, 5. \" \/>\r\nSince we have two points, we will use subscript notation.\r\n<p style=\"text-align: center\">[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(2,3\\right)}\\stackrel{{x}_{2},{y}_{2}}{\\left(7,6\\right)}[\/latex]<\/p>\r\nOn the graph, we counted the rise of [latex]3[\/latex]. The rise can also be found by subtracting the [latex]y\\text{-coordinates}[\/latex] of the points.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{y}_{2}-{y}_{1}\\\\ 6 - 3\\\\ 3\\end{array}[\/latex]<\/p>\r\nWe counted a run of [latex]5[\/latex]. The run can also be found by subtracting the [latex]x\\text{-coordinates}[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{x}_{2}-{x}_{1}\\\\ 7 - 2\\\\ 5\\end{array}[\/latex]<\/p>\r\n\r\n<table id=\"eip-id1168468520883\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We know<\/td>\r\n<td>[latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>So<\/td>\r\n<td>[latex]m={\\Large\\frac{3}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We rewrite the rise and run by putting in the coordinates.<\/td>\r\n<td>[latex]m={\\Large\\frac{6 - 3}{7 - 2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>But [latex]6[\/latex] is the [latex]y[\/latex] -coordinate of the second point, [latex]{y}_{2}[\/latex]\r\n\r\nand [latex]3[\/latex] is the [latex]y[\/latex] -coordinate of the first point [latex]{y}_{1}[\/latex] .\r\n\r\nSo we can rewrite the rise using subscript notation.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{7 - 2}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Also [latex]7[\/latex] is the [latex]x[\/latex] -coordinate of the second point, [latex]{x}_{2}[\/latex]\r\n\r\nand [latex]2[\/latex] is the [latex]x[\/latex] -coordinate of the first point [latex]{x}_{2}[\/latex] .\r\n\r\nSo we rewrite the run using subscript notation.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe\u2019ve shown that [latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex] is really another version of [latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]. We can use this formula to find the slope of a line when we have two points on the line.\r\n<div class=\"textbox shaded\">\r\n<h3>Slope Formula<\/h3>\r\nThe slope of the line between two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] is\r\n\r\n[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]\r\n\r\nSay the formula to yourself to help you remember it:\r\n<p style=\"text-align: center\">[latex]\\text{Slope is }y\\text{ of the second point minus }y\\text{ of the first point}[\/latex]\r\n[latex]\\text{over}[\/latex]\r\n[latex]x\\text{ of the second point minus }x\\text{ of the first point.}[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nFind the slope of the line between the points [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(4,5\\right)[\/latex].\r\n\r\nSolution\r\n<table id=\"eip-id1168468461864\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We\u2019ll call [latex]\\left(1,2\\right)[\/latex] point #1 and [latex]\\left(4,5\\right)[\/latex] point #2.<\/td>\r\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(1,2\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(4,5\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the slope formula.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute the values in the slope formula:<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{5 - 2}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{5 - 2}{4 - 1}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify the numerator and the denominator.<\/td>\r\n<td>[latex]m={\\Large\\frac{3}{3}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]m=1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet\u2019s confirm this by counting out the slope on the graph.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224621\/CNX_BMath_Figure_11_04_031.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -1 to 7. Two labeled points are drawn at \" \/>\r\nThe rise is [latex]3[\/latex] and the run is [latex]3[\/latex], so\r\n[latex]\\begin{array}{}\\\\ m=\\frac{\\text{rise}}{\\text{run}}\\hfill \\\\ m={\\Large\\frac{3}{3}}\\hfill \\\\ m=1\\hfill \\end{array}[\/latex]\r\n\r\n<\/div>\r\nIt is important to remember that the slope is the same no matter which order we select the points.\u00a0 Previously, whenever we found the slope by looking at the graph, we always selected our points from left to right so that our run was always a positive value.\u00a0 Now, let's take a look at an example in which we select our points from right to left.\r\n\r\nThe point [latex](0,2)[\/latex] is indicated as Point 1, and [latex](\u22122,6)[\/latex] as Point 2. So you are going to move from Point 1 to Point 2. A triangle is drawn in above the line to help illustrate the rise and run.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064310\/image032.jpg\" alt=\"A line going through Point 1, or (0,2), and Point 2, or (-2,6). The rise is 4 and the run is -2.\" width=\"410\" height=\"396\" \/>\r\n\r\nYou can see from the graph that the rise going from Point 1 to Point 2 is [latex]4[\/latex], because you are moving [latex]4[\/latex] units in a positive direction (up). The run is [latex]\u22122[\/latex], because you are then moving in a negative direction (left) [latex]2[\/latex] units (think of it like running backwards!). Using the slope formula,\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{-2}=-2[\/latex].<\/p>\r\nYou do not need the graph to find the slope. You can just use the coordinates, keeping careful track of which is Point 1 and which is Point 2. Let\u2019s organize the information about the two points:\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Name<\/th>\r\n<th>Ordered Pair<\/th>\r\n<th>Coordinates<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Point 1<\/td>\r\n<td>[latex](0,2)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Point 2<\/td>\r\n<td>[latex](\u22122,6)[\/latex]<\/td>\r\n<td>[latex]\\begin{array}{l}x_{2}=-2\\\\y_{2}=6\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe slope, [latex]m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{6-2}{-2-0}=\\frac{4}{-2}=-2[\/latex]. The slope of the line, <i>m<\/i>, is [latex]\u22122[\/latex].\r\n\r\nRemember, it doesn\u2019t matter which point is designated as Point 1 and which is Point 2. You could have called [latex](\u22122,6)[\/latex] Point 1, and [latex](0,2)[\/latex] Point 2. In that\u00a0case, putting the coordinates into the slope formula produces the equation [latex]m=\\frac{2-6}{0-\\left(-2\\right)}=\\frac{-4}{2}=-2[\/latex]. Once again, the slope is [latex]m=-2[\/latex]. That\u2019s the same slope as before. The important thing is to be consistent when you subtract: you must always subtract in the same order [latex]\\left(y_{2}-y_{1}\\right)[\/latex]<sub>\u00a0<\/sub>and [latex]\\left(x_{2}-x_{1}\\right)[\/latex].\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]147021[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nFind the slope of the line through the points [latex]\\left(-2,-3\\right)[\/latex] and [latex]\\left(-7,4\\right)[\/latex].\r\n[reveal-answer q=\"265532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265532\"]\r\n\r\nSolution\r\n<table id=\"eip-id1168469701671\" class=\"unnumbered unstyled\" summary=\".\">\r\n<tbody>\r\n<tr>\r\n<td>We\u2019ll call [latex]\\left(-2,-3\\right)[\/latex] point #1 and [latex]\\left(-7,4\\right)[\/latex] point #2.<\/td>\r\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(-2,-3\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(-7,4\\right)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Use the slope formula.<\/td>\r\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Substitute the values<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\r\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{-7-\\left(-2\\right)}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]m={\\Large\\frac{7}{-5}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]m=-{\\Large\\frac{7}{5}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nLet\u2019s confirm this on the graph shown.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224622\/CNX_BMath_Figure_11_04_032.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -8 to 2. The y-axis runs from -6 to 5. Two unlabeled points are drawn at \" \/>\r\n[latex]\\begin{array}{}\\\\ \\\\ \\\\ m=\\frac{\\text{rise}}{\\text{run}}\\\\ m={\\Large\\frac{-7}{5}}\\\\ m=-{\\Large\\frac{7}{5}}\\end{array}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]147022[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](4,2)[\/latex] and [latex](5,5)[\/latex]?\r\n\r\n[reveal-answer q=\"666697\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666697\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=4\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\left(4,2\\right)=\\text{Point }1,\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{2}=5\\\\y_{2}=5\\end{array}[\/latex]<\/td>\r\n<td>[latex]\\left(5,5\\right)=\\text{Point }2,\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{5-2}{5-4}=\\frac{3}{1}\\\\\\\\m=3\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is [latex]3[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nThe example below shows the solution when you reverse the order of the points, calling [latex](5,5)[\/latex] Point 1 and [latex](4,2)[\/latex] Point 2.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?\r\n\r\n[reveal-answer q=\"76013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"76013\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=5\\\\y_{1}=5\\end{array}[\/latex]<\/td>\r\n<td>[latex](5,5)=\\text{Point }1[\/latex], [latex]\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](4,2)=\\text{Point }2[\/latex], [latex]\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{{x_2}-x_{1}}\\\\\\\\m=\\frac{2-5}{4-5}=\\frac{-3}{-1}=3\\\\\\\\m=3\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is [latex]3[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that regardless of which ordered pair is named Point 1 and which is named Point 2, the slope is still [latex]3[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nWhat is the slope of the line that contains the points [latex](3,-6.25)[\/latex] and [latex](-1,8.5)[\/latex]?\r\n\r\n[reveal-answer q=\"291649\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"291649\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\begin{array}{l}x_{1}=3\\\\y_{1}=-6.25\\end{array}[\/latex]<\/td>\r\n<td>[latex](3,-6.25)=\\text{Point }1[\/latex], [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-1\\\\{{y}_{2}}=8.5\\end{array}[\/latex]<\/td>\r\n<td>[latex](-1,8.5)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{8.5-(-6.25)}{-1-3}\\\\\\\\m=\\frac{14.75}{-4}\\\\\\\\m=-3.6875\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is [latex]-3.6875[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nWatch these videos to see more examples of how to determine slope given two points on a line.\r\n\r\nhttps:\/\/youtu.be\/ZW7rQa8SJSU\r\n\r\nhttps:\/\/youtu.be\/6qONExlVGgc\r\n<h2 id=\"Finding Slopes of Horizontal and Vertical Lines\">Finding the Slopes of Horizontal and Vertical Lines<\/h2>\r\nNow, let's revisit horizontal and vertical lines.\u00a0 So far in this section, we have considered lines that run \u201cuphill\u201d or \u201cdownhill.\u201d Their slopes are always positive or negative numbers. But what about horizontal and vertical lines? Can we still use the slope formula to calculate their slopes?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064316\/image040.jpg\" alt=\"The line y=3 crosses through the point (-3,3); the point (0,3); the point (2,3); and the point (5,3).\" width=\"335\" height=\"324\" \/>\r\n\r\nConsider the line above.\u00a0 We learned in the previous section that because it is horizontal, its slope is [latex]0[\/latex]. You can also use the slope formula with two points on this horizontal line to calculate the slope of this horizontal line. Using [latex](\u22123,3)[\/latex] as Point 1 and (2, 3) as Point 2, you get:\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-3}{2-\\left(-3\\right)}=\\frac{0}{5}=0\\end{array}[\/latex]<\/p>\r\nThe slope of this horizontal line is [latex]0[\/latex].\r\n\r\nLet\u2019s consider any horizontal line. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. So, when you apply the slope formula, the numerator will always be [latex]0[\/latex]. Zero divided by any non-zero number is [latex]0[\/latex], so the slope of any horizontal line is always [latex]0[\/latex].\r\n\r\nThe equation for the horizontal line [latex]y=3[\/latex]\u00a0is telling you that no matter which two points you choose on this line, the <i>y-<\/i>coordinate will always be [latex]3[\/latex].\r\n\r\nHow about vertical lines? In their case, no matter which two points you choose, they will always have the same <i>x<\/i>-coordinate. The equation for this line is [latex]x=2[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064317\/image041.jpg\" alt=\"The line x=2 runs through the point (2,-2), the point (2,1), the point (2,3), and the point (2,4).\" width=\"387\" height=\"374\" \/>\r\n\r\nSo, what happens when you use the slope formula with two points on this vertical line to calculate the slope? Using [latex](2,1)[\/latex] as Point 1 and [latex](2,3)[\/latex] as Point 2, you get:\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-1}{2-2}=\\frac{2}{0}\\end{array}[\/latex]<\/p>\r\nBut division by zero has no meaning for the set of real numbers. Because of this fact, it is said that the slope of this vertical line is undefined. This is true for all vertical lines\u2014they all have a slope that is undefined.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](3,2)[\/latex] and [latex](\u22128,2)[\/latex]?\r\n\r\n[reveal-answer q=\"111566\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111566\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{1}}=3\\\\{{y}_{1}}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](3,2)=\\text{Point }1[\/latex], [latex] \\displaystyle \\left(x_{1},x_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-8\\\\\\\\{{y}_{2}}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](\u22128,2)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\\\frac{(2)-(2)}{(-8)-(3)}=\\frac{0}{-11}=0\\\\\\\\m=0\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is [latex]0[\/latex], so the line is horizontal.[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]76720[\/ohm_question]\r\n\r\n[ohm_question]195555[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nhttps:\/\/youtu.be\/zoLM3rxzndo\r\n<h2><\/h2>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the slope formula to find the slope of a line between two points<\/li>\n<li>Find the slope of horizontal and vertical lines<\/li>\n<\/ul>\n<\/div>\n<p>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.<\/p>\n<p>Before we get to it, we need to introduce some new algebraic notation. We have seen that an ordered pair [latex]\\left(x,y\\right)[\/latex] gives the coordinates of a point. But when we work with slopes, we use two points. How can the same symbol [latex]\\left(x,y\\right)[\/latex] be used to represent two different points?<\/p>\n<p>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.<\/p>\n<p>We will use [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] to identify the first point and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] to identify the second point. (If we had more than two points, we could use [latex]\\left({x}_{3},{y}_{3}\\right),\\left({x}_{4},{y}_{4}\\right)[\/latex], and so on.)<\/p>\n<h2>The Slope Formula<\/h2>\n<p>You\u2019ve seen that you can find the slope of a line on a graph by measuring the rise and the run. You can also find the slope of a straight line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an\u00a0[latex]x[\/latex]-value and a [latex]y[\/latex]-value, written as an ordered pair [latex](x, y)[\/latex]. The [latex]x[\/latex] value tells you where a point is horizontally. The [latex]y[\/latex] value tells you where the point is vertically.<\/p>\n<p>Consider two points on a line\u2014Point 1 and Point 2. Point 1 has coordinates [latex]\\left(x_{1},y_{1}\\right)[\/latex]\u00a0and Point 2 has coordinates [latex]\\left(x_{2},y_{2}\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064308\/image031.jpg\" alt=\"A line with its rise and run. The first point on the line is labeled Point 1, or (x1, y1). The second point on the line is labeled Point 2, or (x2,y2). The rise is (y2 minus y1). The run is (x2 minus X1).\" width=\"416\" height=\"401\" \/><\/p>\n<p>The rise is the vertical distance between the two points, which is the difference between their \u00a0[latex]y[\/latex]-coordinates. That makes the rise [latex]\\left(y_{2}-y_{1}\\right)[\/latex]. The run between these two points is the difference in the [latex]x[\/latex]-coordinates, or [latex]\\left(x_{2}-x_{1}\\right)[\/latex].<\/p>\n<p>So, [latex]\\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex] or [latex]\\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex].<\/p>\n<p>To see how the rise and run relate to the coordinates of the two points, let\u2019s take another look at the slope of the line between the points [latex]\\left(2,3\\right)[\/latex] and [latex]\\left(7,6\\right)[\/latex] below.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224620\/CNX_BMath_Figure_11_04_030.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from 0 to 7. The y-axis runs from 0 to 7. A line runs through the labeled points 2, 3 and 7, 6. A line segment runs from the point 2, 3 to the unlabeled point 2, 6. It is labeled y sub 2 minus y sub 1, 6 minus 3, 3. A line segment runs from the point 7, 6 to the unlabeled point 2, 6. It os labeled x sub 2 minus x sub 1, 7 minus 2, 5.\" \/><br \/>\nSince we have two points, we will use subscript notation.<\/p>\n<p style=\"text-align: center\">[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(2,3\\right)}\\stackrel{{x}_{2},{y}_{2}}{\\left(7,6\\right)}[\/latex]<\/p>\n<p>On the graph, we counted the rise of [latex]3[\/latex]. The rise can also be found by subtracting the [latex]y\\text{-coordinates}[\/latex] of the points.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{y}_{2}-{y}_{1}\\\\ 6 - 3\\\\ 3\\end{array}[\/latex]<\/p>\n<p>We counted a run of [latex]5[\/latex]. The run can also be found by subtracting the [latex]x\\text{-coordinates}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}{x}_{2}-{x}_{1}\\\\ 7 - 2\\\\ 5\\end{array}[\/latex]<\/p>\n<table id=\"eip-id1168468520883\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We know<\/td>\n<td>[latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>So<\/td>\n<td>[latex]m={\\Large\\frac{3}{5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>We rewrite the rise and run by putting in the coordinates.<\/td>\n<td>[latex]m={\\Large\\frac{6 - 3}{7 - 2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>But [latex]6[\/latex] is the [latex]y[\/latex] -coordinate of the second point, [latex]{y}_{2}[\/latex]<\/p>\n<p>and [latex]3[\/latex] is the [latex]y[\/latex] -coordinate of the first point [latex]{y}_{1}[\/latex] .<\/p>\n<p>So we can rewrite the rise using subscript notation.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{7 - 2}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Also [latex]7[\/latex] is the [latex]x[\/latex] -coordinate of the second point, [latex]{x}_{2}[\/latex]<\/p>\n<p>and [latex]2[\/latex] is the [latex]x[\/latex] -coordinate of the first point [latex]{x}_{2}[\/latex] .<\/p>\n<p>So we rewrite the run using subscript notation.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We\u2019ve shown that [latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex] is really another version of [latex]m={\\Large\\frac{\\text{rise}}{\\text{run}}}[\/latex]. We can use this formula to find the slope of a line when we have two points on the line.<\/p>\n<div class=\"textbox shaded\">\n<h3>Slope Formula<\/h3>\n<p>The slope of the line between two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] is<\/p>\n<p>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/p>\n<p>Say the formula to yourself to help you remember it:<\/p>\n<p style=\"text-align: center\">[latex]\\text{Slope is }y\\text{ of the second point minus }y\\text{ of the first point}[\/latex]<br \/>\n[latex]\\text{over}[\/latex]<br \/>\n[latex]x\\text{ of the second point minus }x\\text{ of the first point.}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Find the slope of the line between the points [latex]\\left(1,2\\right)[\/latex] and [latex]\\left(4,5\\right)[\/latex].<\/p>\n<p>Solution<\/p>\n<table id=\"eip-id1168468461864\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We\u2019ll call [latex]\\left(1,2\\right)[\/latex] point #1 and [latex]\\left(4,5\\right)[\/latex] point #2.<\/td>\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(1,2\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(4,5\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the slope formula.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute the values in the slope formula:<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{5 - 2}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{5 - 2}{4 - 1}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify the numerator and the denominator.<\/td>\n<td>[latex]m={\\Large\\frac{3}{3}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]m=1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let\u2019s confirm this by counting out the slope on the graph.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224621\/CNX_BMath_Figure_11_04_031.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -1 to 7. The y-axis runs from -1 to 7. Two labeled points are drawn at\" \/><br \/>\nThe rise is [latex]3[\/latex] and the run is [latex]3[\/latex], so<br \/>\n[latex]\\begin{array}{}\\\\ m=\\frac{\\text{rise}}{\\text{run}}\\hfill \\\\ m={\\Large\\frac{3}{3}}\\hfill \\\\ m=1\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<p>It is important to remember that the slope is the same no matter which order we select the points.\u00a0 Previously, whenever we found the slope by looking at the graph, we always selected our points from left to right so that our run was always a positive value.\u00a0 Now, let&#8217;s take a look at an example in which we select our points from right to left.<\/p>\n<p>The point [latex](0,2)[\/latex] is indicated as Point 1, and [latex](\u22122,6)[\/latex] as Point 2. So you are going to move from Point 1 to Point 2. A triangle is drawn in above the line to help illustrate the rise and run.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064310\/image032.jpg\" alt=\"A line going through Point 1, or (0,2), and Point 2, or (-2,6). The rise is 4 and the run is -2.\" width=\"410\" height=\"396\" \/><\/p>\n<p>You can see from the graph that the rise going from Point 1 to Point 2 is [latex]4[\/latex], because you are moving [latex]4[\/latex] units in a positive direction (up). The run is [latex]\u22122[\/latex], because you are then moving in a negative direction (left) [latex]2[\/latex] units (think of it like running backwards!). Using the slope formula,<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{-2}=-2[\/latex].<\/p>\n<p>You do not need the graph to find the slope. You can just use the coordinates, keeping careful track of which is Point 1 and which is Point 2. Let\u2019s organize the information about the two points:<\/p>\n<table>\n<thead>\n<tr>\n<th>Name<\/th>\n<th>Ordered Pair<\/th>\n<th>Coordinates<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Point 1<\/td>\n<td>[latex](0,2)[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=2\\end{array}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Point 2<\/td>\n<td>[latex](\u22122,6)[\/latex]<\/td>\n<td>[latex]\\begin{array}{l}x_{2}=-2\\\\y_{2}=6\\end{array}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The slope, [latex]m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{6-2}{-2-0}=\\frac{4}{-2}=-2[\/latex]. The slope of the line, <i>m<\/i>, is [latex]\u22122[\/latex].<\/p>\n<p>Remember, it doesn\u2019t matter which point is designated as Point 1 and which is Point 2. You could have called [latex](\u22122,6)[\/latex] Point 1, and [latex](0,2)[\/latex] Point 2. In that\u00a0case, putting the coordinates into the slope formula produces the equation [latex]m=\\frac{2-6}{0-\\left(-2\\right)}=\\frac{-4}{2}=-2[\/latex]. Once again, the slope is [latex]m=-2[\/latex]. That\u2019s the same slope as before. The important thing is to be consistent when you subtract: you must always subtract in the same order [latex]\\left(y_{2}-y_{1}\\right)[\/latex]<sub>\u00a0<\/sub>and [latex]\\left(x_{2}-x_{1}\\right)[\/latex].<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm147021\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=147021&theme=oea&iframe_resize_id=ohm147021&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Find the slope of the line through the points [latex]\\left(-2,-3\\right)[\/latex] and [latex]\\left(-7,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265532\">Show Solution<\/span><\/p>\n<div id=\"q265532\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solution<\/p>\n<table id=\"eip-id1168469701671\" class=\"unnumbered unstyled\" summary=\".\">\n<tbody>\n<tr>\n<td>We\u2019ll call [latex]\\left(-2,-3\\right)[\/latex] point #1 and [latex]\\left(-7,4\\right)[\/latex] point #2.<\/td>\n<td>[latex]\\stackrel{{x}_{1},{y}_{1}}{\\left(-2,-3\\right)}\\text{and}\\stackrel{{x}_{2},{y}_{2}}{\\left(-7,4\\right)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Use the slope formula.<\/td>\n<td>[latex]m={\\Large\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Substitute the values<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td>[latex]y[\/latex] of the second point minus [latex]y[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{{x}_{2}-{x}_{1}}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]x[\/latex] of the second point minus [latex]x[\/latex] of the first point<\/td>\n<td>[latex]m={\\Large\\frac{4-\\left(-3\\right)}{-7-\\left(-2\\right)}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]m={\\Large\\frac{7}{-5}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td>[latex]m=-{\\Large\\frac{7}{5}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Let\u2019s confirm this on the graph shown.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/25224622\/CNX_BMath_Figure_11_04_032.png\" alt=\"The graph shows the x y-coordinate plane. The x-axis runs from -8 to 2. The y-axis runs from -6 to 5. Two unlabeled points are drawn at\" \/><br \/>\n[latex]\\begin{array}{}\\\\ \\\\ \\\\ m=\\frac{\\text{rise}}{\\text{run}}\\\\ m={\\Large\\frac{-7}{5}}\\\\ m=-{\\Large\\frac{7}{5}}\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm147022\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=147022&theme=oea&iframe_resize_id=ohm147022&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](4,2)[\/latex] and [latex](5,5)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q666697\">Show Solution<\/span><\/p>\n<div id=\"q666697\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=4\\\\y_{1}=2\\end{array}[\/latex]<\/td>\n<td>[latex]\\left(4,2\\right)=\\text{Point }1,\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}x_{2}=5\\\\y_{2}=5\\end{array}[\/latex]<\/td>\n<td>[latex]\\left(5,5\\right)=\\text{Point }2,\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{5-2}{5-4}=\\frac{3}{1}\\\\\\\\m=3\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is [latex]3[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<p>The example below shows the solution when you reverse the order of the points, calling [latex](5,5)[\/latex] Point 1 and [latex](4,2)[\/latex] Point 2.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q76013\">Show Solution<\/span><\/p>\n<div id=\"q76013\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=5\\\\y_{1}=5\\end{array}[\/latex]<\/td>\n<td>[latex](5,5)=\\text{Point }1[\/latex], [latex]\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=2\\end{array}[\/latex]<\/td>\n<td>[latex](4,2)=\\text{Point }2[\/latex], [latex]\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\begin{array}{l}m=\\frac{y_{2}-y_{1}}{{x_2}-x_{1}}\\\\\\\\m=\\frac{2-5}{4-5}=\\frac{-3}{-1}=3\\\\\\\\m=3\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is [latex]3[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<p>Notice that regardless of which ordered pair is named Point 1 and which is named Point 2, the slope is still [latex]3[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>What is the slope of the line that contains the points [latex](3,-6.25)[\/latex] and [latex](-1,8.5)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q291649\">Show Solution<\/span><\/p>\n<div id=\"q291649\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\begin{array}{l}x_{1}=3\\\\y_{1}=-6.25\\end{array}[\/latex]<\/td>\n<td>[latex](3,-6.25)=\\text{Point }1[\/latex], [latex]\\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{2}}=-1\\\\{{y}_{2}}=8.5\\end{array}[\/latex]<\/td>\n<td>[latex](-1,8.5)=\\text{Point }2[\/latex], [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{8.5-(-6.25)}{-1-3}\\\\\\\\m=\\frac{14.75}{-4}\\\\\\\\m=-3.6875\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is [latex]-3.6875[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>Watch these videos to see more examples of how to determine slope given two points on a line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Determine the Slope a Line Given Two Points on a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ZW7rQa8SJSU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 2:  Determine the Slope a Line Given Two Points on a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6qONExlVGgc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2 id=\"Finding Slopes of Horizontal and Vertical Lines\">Finding the Slopes of Horizontal and Vertical Lines<\/h2>\n<p>Now, let&#8217;s revisit horizontal and vertical lines.\u00a0 So far in this section, we have considered lines that run \u201cuphill\u201d or \u201cdownhill.\u201d Their slopes are always positive or negative numbers. But what about horizontal and vertical lines? Can we still use the slope formula to calculate their slopes?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064316\/image040.jpg\" alt=\"The line y=3 crosses through the point (-3,3); the point (0,3); the point (2,3); and the point (5,3).\" width=\"335\" height=\"324\" \/><\/p>\n<p>Consider the line above.\u00a0 We learned in the previous section that because it is horizontal, its slope is [latex]0[\/latex]. You can also use the slope formula with two points on this horizontal line to calculate the slope of this horizontal line. Using [latex](\u22123,3)[\/latex] as Point 1 and (2, 3) as Point 2, you get:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-3}{2-\\left(-3\\right)}=\\frac{0}{5}=0\\end{array}[\/latex]<\/p>\n<p>The slope of this horizontal line is [latex]0[\/latex].<\/p>\n<p>Let\u2019s consider any horizontal line. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. So, when you apply the slope formula, the numerator will always be [latex]0[\/latex]. Zero divided by any non-zero number is [latex]0[\/latex], so the slope of any horizontal line is always [latex]0[\/latex].<\/p>\n<p>The equation for the horizontal line [latex]y=3[\/latex]\u00a0is telling you that no matter which two points you choose on this line, the <i>y-<\/i>coordinate will always be [latex]3[\/latex].<\/p>\n<p>How about vertical lines? In their case, no matter which two points you choose, they will always have the same <i>x<\/i>-coordinate. The equation for this line is [latex]x=2[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064317\/image041.jpg\" alt=\"The line x=2 runs through the point (2,-2), the point (2,1), the point (2,3), and the point (2,4).\" width=\"387\" height=\"374\" \/><\/p>\n<p>So, what happens when you use the slope formula with two points on this vertical line to calculate the slope? Using [latex](2,1)[\/latex] as Point 1 and [latex](2,3)[\/latex] as Point 2, you get:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-1}{2-2}=\\frac{2}{0}\\end{array}[\/latex]<\/p>\n<p>But division by zero has no meaning for the set of real numbers. Because of this fact, it is said that the slope of this vertical line is undefined. This is true for all vertical lines\u2014they all have a slope that is undefined.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](3,2)[\/latex] and [latex](\u22128,2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q111566\">Show Solution<\/span><\/p>\n<div id=\"q111566\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{1}}=3\\\\{{y}_{1}}=2\\end{array}[\/latex]<\/td>\n<td>[latex](3,2)=\\text{Point }1[\/latex], [latex]\\displaystyle \\left(x_{1},x_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{2}}=-8\\\\\\\\{{y}_{2}}=2\\end{array}[\/latex]<\/td>\n<td>[latex](\u22128,2)=\\text{Point }2[\/latex], [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\\\frac{(2)-(2)}{(-8)-(3)}=\\frac{0}{-11}=0\\\\\\\\m=0\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is [latex]0[\/latex], so the line is horizontal.<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm76720\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=76720&theme=oea&iframe_resize_id=ohm76720&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm195555\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=195555&theme=oea&iframe_resize_id=ohm195555&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Determine the Slope a Line Given Two Points on a Horizontal and Vertical Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zoLM3rxzndo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><\/h2>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-10698\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Question ID 147022, 147021, 147020. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Determine the Slope a Line Given Two Points on a Line. <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6qONExlVGgc\">https:\/\/youtu.be\/6qONExlVGgc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Prealgebra. <strong>Provided by<\/strong>: OpenStax. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"},{\"type\":\"original\",\"description\":\"Question ID 147022, 147021, 147020\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Determine the Slope a Line Given Two Points on a Line\",\"author\":\"James Sousa (mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/6qONExlVGgc\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"012082040f0848959b64162572e7cb1a, 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