{"id":939,"date":"2016-10-20T20:53:26","date_gmt":"2016-10-20T20:53:26","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=939"},"modified":"2017-04-10T20:20:04","modified_gmt":"2017-04-10T20:20:04","slug":"parallel-and-perpendicular-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/parallel-and-perpendicular-lines\/","title":{"raw":"Parallel and Perpendicular Lines","rendered":"Parallel and Perpendicular Lines"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Determine whether two lines are parallel or perpendicular<\/li>\r\n \t<li>Find the equations of parallel and perpendicular lines<\/li>\r\n \t<li>Write the equations of lines that are parallel or perpendicular to a given line<\/li>\r\n<\/ul>\r\n<\/div>\r\nParallel lines have the same slope and different <em>y-<\/em>intercepts. Lines that are <strong>parallel<\/strong> to each other will never intersect. For example, the figure below\u00a0shows the graphs of various lines with the same slope, [latex]m=2[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232102\/CNX_CAT_Figure_02_02_004.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.\" width=\"487\" height=\"593\" data-media-type=\"image\/jpg\" \/> Parallel lines[\/caption]\r\n\r\nAll of the lines shown in the graph are parallel because they have the same slope and different <em>y-<\/em>intercepts.\r\n\r\nLines that are <strong>perpendicular<\/strong> intersect to form a [latex]{90}^{\\circ }[\/latex] -angle. The slope of one line is the negative <strong>reciprocal<\/strong> of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\\cdot {m}_{2}=-1[\/latex]. For example, the figure above\u00a0shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\\frac{1}{3}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{m}_{1}\\cdot {m}_{2}=-1\\hfill \\\\ \\text{ }3\\cdot \\left(-\\frac{1}{3}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232106\/CNX_CAT_Figure_02_02_005.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x\/3 minus 2. Their intersection is marked by a box to show that it is a right angle.\" width=\"487\" height=\"329\" data-media-type=\"image\/jpg\" \/> Perpendicular lines[\/caption]\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither<\/h3>\r\nGraph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[\/latex] and [latex]3x - 4y=8[\/latex].\r\n[reveal-answer q=\"130903\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"130903\"]\r\n\r\nThe first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.\r\n\r\nFirst equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=-4x+3\\hfill \\\\ y=-\\frac{4}{3}x+1\\hfill \\end{array}[\/latex]<\/p>\r\nSecond equation:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x - 4y=8\\hfill \\\\ -4y=-3x+8\\hfill \\\\ y=\\frac{3}{4}x - 2\\hfill \\end{array}[\/latex]<\/p>\r\nSee the graph of both lines in the graph below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232109\/CNX_CAT_Figure_02_02_006.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x\/3 plus 1 and y = 3 times x\/4 minus 2. A box is placed at the intersection to note that it forms a right angle.\" width=\"487\" height=\"366\" data-media-type=\"image\/jpg\" \/>\r\n\r\nFrom the graph, we can see that the lines appear perpendicular, but we must compare the slopes.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=-\\frac{4}{3}\\hfill \\\\ {m}_{2}=\\frac{3}{4}\\hfill \\\\ {m}_{1}\\cdot {m}_{2}=\\left(-\\frac{4}{3}\\right)\\left(\\frac{3}{4}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\r\nThe slopes are negative reciprocals of each other, confirming that the lines are perpendicular.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It now<\/h3>\r\nGraph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[\/latex] and [latex]2y=x+4[\/latex].\r\n\r\n[reveal-answer q=\"727314\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"727314\"]\r\n\r\nParallel lines: equations are written in slope-intercept form.\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200328\/CNX_CAT_Figure_02_02_007.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x\/2 plus 5 and y = x\/2 plus 2. The lines do not cross.\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110960&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"750\"><\/iframe>\r\n\r\n<\/div>\r\nIf we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.\r\n<h2>Writing Equations of Parallel Lines<\/h2>\r\nSuppose for example, we are given the following equation.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\\latex]<\/p>\r\nWe know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{cases}[\/latex]<\/p>\r\nSuppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{ }y=3x+4\\hfill \\end{cases}[\/latex]<\/p>\r\nSo [latex]g\\left(x\\right)=3x+4[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1[\/latex] and passes through the point (1, 7).\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the function.<\/li>\r\n \t<li>Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\r\nFind a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"672987\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"672987\"]\r\nThe slope of the given line is 3. If we choose the slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and <em>f<\/em>(<em>x<\/em>) = 0 into the slope-intercept form to find the <em>y-<\/em>intercept.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{ }0=3\\left(3\\right)+b\\hfill \\\\ \\text{ }b=-9\\hfill \\end{cases}[\/latex]<\/p>\r\nThe line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" data-media-type=\"image\/jpg\" \/>\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<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110970&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\nCheck your work with Desmos.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Writing Equations of Perpendicular Lines<\/h2>\r\nWe can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/p>\r\nThe slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{cases}[\/latex]<\/p>\r\nAs before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{cases}[\/latex]<\/p>\r\nThe equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].\r\n\r\nSo [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong>\r\n\r\n<em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the function.<\/li>\r\n \t<li>Determine the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t<li>Solve for <em>b<\/em>.<\/li>\r\n \t<li>Write the equation for the line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\r\nFind the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"566588\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"566588\"]\r\n\r\nThe original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}y_{2}=-\\frac{1}{3}x+b\\hfill \\\\ \\text{ }0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{cases}[\/latex]<\/p>\r\nThe line perpendicular to <i>y<\/i>\u00a0that passes through (3, 0) is [latex]y_{2}=-\\frac{1}{3}x+1[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the two lines is shown in the graph below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGiven the line [latex]y=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is\r\n<ol>\r\n \t<li>parallel to\u00a0<em>y<\/em><\/li>\r\n \t<li>perpendicular to\u00a0<em>y<\/em><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"924175\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"924175\"]\r\n<ol>\r\n \t<li><span data-type=\"item\" data-label=\"a\">[latex]y=2x[\/latex]is parallel <\/span><\/li>\r\n \t<li><span data-type=\"item\" data-label=\"b\">[latex]y=-\\frac{1}{2}x[\/latex]is perpendicular<\/span><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110971&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\nCheck your work with Desmos.\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\r\n<ol>\r\n \t<li>Determine the slope of the line passing through the points.<\/li>\r\n \t<li>Find the negative reciprocal of the slope.<\/li>\r\n \t<li>Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\r\nA line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).\r\n\r\n[reveal-answer q=\"119051\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"119051\"]\r\nFrom the two points of the given line, we can calculate the slope of that line.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ =\\frac{-1}{6}\\hfill \\\\ =-\\frac{1}{6}\\hfill \\end{cases}[\/latex]<\/p>\r\nFind the negative reciprocal of the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ =-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ =6\\hfill \\end{cases}[\/latex]<\/p>\r\nWe can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).\r\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{cases}[\/latex]<\/p>\r\nThe equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is\r\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nA line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).\r\n\r\n[reveal-answer q=\"69699\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"69699\"][latex]y=-\\frac{1}{3}x+6[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Writing the Equations of Lines Parallel or Perpendicular to a Given Line<\/h2>\r\nAs we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the <strong>point-slope formula<\/strong> to write the equation of the new line.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation for a line, write the equation of a line parallel or perpendicular to it.<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.<\/li>\r\n \t<li>Use the slope and the given point with the point-slope formula.<\/li>\r\n \t<li>Simplify the line to slope-intercept form and compare the equation to the given line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point<\/h3>\r\nWrite the equation of line parallel to a [latex]5x+3y=1[\/latex] and passing through the point [latex]\\left(3,5\\right)[\/latex].\r\n[reveal-answer q=\"72460\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"72460\"]\r\n\r\nFirst, we will write the equation in slope-intercept form to find the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x+3y=1\\hfill \\\\ 3y=-5x+1\\hfill \\\\ y=-\\frac{5}{3}x+\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe slope is [latex]m=-\\frac{5}{3}[\/latex]. The <em>y-<\/em>intercept is [latex]\\frac{1}{3}[\/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope.\r\n\r\nThe one exception is that if the <em>y-<\/em>intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 5=-\\frac{5}{3}\\left(x - 3\\right)\\hfill \\\\ y - 5=-\\frac{5}{3}x+5\\hfill \\\\ y=-\\frac{5}{3}x+10\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation of the line is [latex]y=-\\frac{5}{3}x+10[\/latex].\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232111\/CNX_CAT_Figure_02_02_008.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 2 to 12 in intervals of 2. Two functions are graphed on the same plot: y = negative 5 times x\/3 plus 1\/3 and y = negative 5 times x\/3 plus 10. The lines do not cross.\" width=\"487\" height=\"329\" data-media-type=\"image\/jpg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the equation of the line parallel to [latex]5x=7+y[\/latex] and passing through the point [latex]\\left(-1,-2\\right)[\/latex].\r\n\r\n[reveal-answer q=\"985300\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"985300\"][latex]y=5x+3[\/latex][\/hidden-answer]\r\n<iframe id=\"mom7\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1436&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point<\/h3>\r\nFind the equation of the line perpendicular to [latex]5x - 3y+4=0\\left(-4,1\\right)[\/latex].\r\n[reveal-answer q=\"408679\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"408679\"]\r\n\r\nThe first step is to write the equation in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x - 3y+4=0\\hfill \\\\ -3y=-5x - 4\\hfill \\\\ y=\\frac{5}{3}x+\\frac{4}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nWe see that the slope is [latex]m=\\frac{5}{3}[\/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal, or [latex]-\\frac{3}{5}[\/latex]. Next, we use the point-slope formula with this new slope and the given point.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=-\\frac{3}{5}\\left(x-\\left(-4\\right)\\right)\\hfill \\\\ y - 1=-\\frac{3}{5}x-\\frac{12}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{12}{5}+\\frac{5}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{7}{5}\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Determine whether two lines are parallel or perpendicular<\/li>\n<li>Find the equations of parallel and perpendicular lines<\/li>\n<li>Write the equations of lines that are parallel or perpendicular to a given line<\/li>\n<\/ul>\n<\/div>\n<p>Parallel lines have the same slope and different <em>y-<\/em>intercepts. Lines that are <strong>parallel<\/strong> to each other will never intersect. For example, the figure below\u00a0shows the graphs of various lines with the same slope, [latex]m=2[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232102\/CNX_CAT_Figure_02_02_004.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7. Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.\" width=\"487\" height=\"593\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Parallel lines<\/p>\n<\/div>\n<p>All of the lines shown in the graph are parallel because they have the same slope and different <em>y-<\/em>intercepts.<\/p>\n<p>Lines that are <strong>perpendicular<\/strong> intersect to form a [latex]{90}^{\\circ }[\/latex] -angle. The slope of one line is the negative <strong>reciprocal<\/strong> of the other. We can show that two lines are perpendicular if the product of the two slopes is [latex]-1:{m}_{1}\\cdot {m}_{2}=-1[\/latex]. For example, the figure above\u00a0shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of [latex]-\\frac{1}{3}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }{m}_{1}\\cdot {m}_{2}=-1\\hfill \\\\ \\text{ }3\\cdot \\left(-\\frac{1}{3}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232106\/CNX_CAT_Figure_02_02_005.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5. Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x\/3 minus 2. Their intersection is marked by a box to show that it is a right angle.\" width=\"487\" height=\"329\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\">Perpendicular lines<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither<\/h3>\n<p>Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: [latex]3y=-4x+3[\/latex] and [latex]3x - 4y=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q130903\">Solution<\/span><\/p>\n<div id=\"q130903\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.<\/p>\n<p>First equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3y=-4x+3\\hfill \\\\ y=-\\frac{4}{3}x+1\\hfill \\end{array}[\/latex]<\/p>\n<p>Second equation:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}3x - 4y=8\\hfill \\\\ -4y=-3x+8\\hfill \\\\ y=\\frac{3}{4}x - 2\\hfill \\end{array}[\/latex]<\/p>\n<p>See the graph of both lines in the graph below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232109\/CNX_CAT_Figure_02_02_006.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 4 to 5 and the y-axis ranging from negative 4 to 4. Two functions are graphed on the same plot: y = negative 4 times x\/3 plus 1 and y = 3 times x\/4 minus 2. A box is placed at the intersection to note that it forms a right angle.\" width=\"487\" height=\"366\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=-\\frac{4}{3}\\hfill \\\\ {m}_{2}=\\frac{3}{4}\\hfill \\\\ {m}_{1}\\cdot {m}_{2}=\\left(-\\frac{4}{3}\\right)\\left(\\frac{3}{4}\\right)=-1\\hfill \\end{array}[\/latex]<\/p>\n<p>The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It now<\/h3>\n<p>Graph the two lines and determine whether they are parallel, perpendicular, or neither: [latex]2y-x=10[\/latex] and [latex]2y=x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q727314\">Solution<\/span><\/p>\n<div id=\"q727314\" class=\"hidden-answer\" style=\"display: none\">\n<p>Parallel lines: equations are written in slope-intercept form.<br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200328\/CNX_CAT_Figure_02_02_007.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 5 to 5 and the y-axis ranging from negative 1 to 6. Two functions are graphed on the same plot: y = x\/2 plus 5 and y = x\/2 plus 2. The lines do not cross.\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110960&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"750\"><\/iframe><\/p>\n<\/div>\n<p>If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n<h2>Writing Equations of Parallel Lines<\/h2>\n<p>Suppose for example, we are given the following equation.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\\latex]<\/p>\n<p>  We know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).  <\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{cases}[\/latex]<\/p>\n<p>Suppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{ }y=3x+4\\hfill \\end{cases}[\/latex]<\/p>\n<p>So [latex]g\\left(x\\right)=3x+4[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1[\/latex] and passes through the point (1, 7).<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\n<ol>\n<li>Find the slope of the function.<\/li>\n<li>Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\n<p>Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q672987\">Solution<\/span><\/p>\n<div id=\"q672987\" class=\"hidden-answer\" style=\"display: none\">\nThe slope of the given line is 3. If we choose the slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and <em>f<\/em>(<em>x<\/em>) = 0 into the slope-intercept form to find the <em>y-<\/em>intercept.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{ }0=3\\left(3\\right)+b\\hfill \\\\ \\text{ }b=-9\\hfill \\end{cases}[\/latex]<\/p>\n<p>The line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can confirm that the two lines are parallel by graphing them. The graph below\u00a0shows that the two lines will never intersect.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184416\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" data-media-type=\"image\/jpg\" \/><\/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=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110970&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><br \/>\nCheck your work with Desmos.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Writing Equations of Perpendicular Lines<\/h2>\n<p>We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/p>\n<p>The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{cases}[\/latex]<\/p>\n<p>As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{cases}[\/latex]<\/p>\n<p>The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/p>\n<p>So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n<p><em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\n<ol>\n<li>Find the slope of the function.<\/li>\n<li>Determine the negative reciprocal of the slope.<\/li>\n<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\n<li>Solve for <em>b<\/em>.<\/li>\n<li>Write the equation for the line.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\n<p>Find the equation of a line perpendicular to [latex]y=3x+3[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q566588\">Solution<\/span><\/p>\n<div id=\"q566588\" class=\"hidden-answer\" style=\"display: none\">\n<p>The original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}y_{2}=-\\frac{1}{3}x+b\\hfill \\\\ \\text{ }0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{cases}[\/latex]<\/p>\n<p>The line perpendicular to <i>y<\/i>\u00a0that passes through (3, 0) is [latex]y_{2}=-\\frac{1}{3}x+1[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the two lines is shown in the graph below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184419\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Given the line [latex]y=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\n<ol>\n<li>parallel to\u00a0<em>y<\/em><\/li>\n<li>perpendicular to\u00a0<em>y<\/em><\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924175\">Solution<\/span><\/p>\n<div id=\"q924175\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li><span data-type=\"item\" data-label=\"a\">[latex]y=2x[\/latex]is parallel <\/span><\/li>\n<li><span data-type=\"item\" data-label=\"b\">[latex]y=-\\frac{1}{2}x[\/latex]is perpendicular<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110971&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><br \/>\nCheck your work with Desmos.<br \/>\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\n<ol>\n<li>Determine the slope of the line passing through the points.<\/li>\n<li>Find the negative reciprocal of the slope.<\/li>\n<li>Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\n<p>A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q119051\">Solution<\/span><\/p>\n<div id=\"q119051\" class=\"hidden-answer\" style=\"display: none\">\nFrom the two points of the given line, we can calculate the slope of that line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ =\\frac{-1}{6}\\hfill \\\\ =-\\frac{1}{6}\\hfill \\end{cases}[\/latex]<\/p>\n<p>Find the negative reciprocal of the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ =-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ =6\\hfill \\end{cases}[\/latex]<\/p>\n<p>We can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{cases}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{cases}[\/latex]<\/p>\n<p>The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is<\/p>\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q69699\">Solution<\/span><\/p>\n<div id=\"q69699\" class=\"hidden-answer\" style=\"display: none\">[latex]y=-\\frac{1}{3}x+6[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Writing the Equations of Lines Parallel or Perpendicular to a Given Line<\/h2>\n<p>As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the <strong>point-slope formula<\/strong> to write the equation of the new line.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an equation for a line, write the equation of a line parallel or perpendicular to it.<\/h3>\n<ol>\n<li>Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.<\/li>\n<li>Use the slope and the given point with the point-slope formula.<\/li>\n<li>Simplify the line to slope-intercept form and compare the equation to the given line.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point<\/h3>\n<p>Write the equation of line parallel to a [latex]5x+3y=1[\/latex] and passing through the point [latex]\\left(3,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q72460\">Solution<\/span><\/p>\n<div id=\"q72460\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we will write the equation in slope-intercept form to find the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x+3y=1\\hfill \\\\ 3y=-5x+1\\hfill \\\\ y=-\\frac{5}{3}x+\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>The slope is [latex]m=-\\frac{5}{3}[\/latex]. The <em>y-<\/em>intercept is [latex]\\frac{1}{3}[\/latex], but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope.<\/p>\n<p>The one exception is that if the <em>y-<\/em>intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 5=-\\frac{5}{3}\\left(x - 3\\right)\\hfill \\\\ y - 5=-\\frac{5}{3}x+5\\hfill \\\\ y=-\\frac{5}{3}x+10\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation of the line is [latex]y=-\\frac{5}{3}x+10[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/17232111\/CNX_CAT_Figure_02_02_008.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 2 to 12 in intervals of 2. Two functions are graphed on the same plot: y = negative 5 times x\/3 plus 1\/3 and y = negative 5 times x\/3 plus 10. The lines do not cross.\" width=\"487\" height=\"329\" data-media-type=\"image\/jpg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the equation of the line parallel to [latex]5x=7+y[\/latex] and passing through the point [latex]\\left(-1,-2\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q985300\">Solution<\/span><\/p>\n<div id=\"q985300\" class=\"hidden-answer\" style=\"display: none\">[latex]y=5x+3[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom7\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=1436&amp;theme=oea&amp;iframe_resize_id=mom7\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point<\/h3>\n<p>Find the equation of the line perpendicular to [latex]5x - 3y+4=0\\left(-4,1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q408679\">Solution<\/span><\/p>\n<div id=\"q408679\" class=\"hidden-answer\" style=\"display: none\">\n<p>The first step is to write the equation in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}5x - 3y+4=0\\hfill \\\\ -3y=-5x - 4\\hfill \\\\ y=\\frac{5}{3}x+\\frac{4}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>We see that the slope is [latex]m=\\frac{5}{3}[\/latex]. This means that the slope of the line perpendicular to the given line is the negative reciprocal, or [latex]-\\frac{3}{5}[\/latex]. Next, we use the point-slope formula with this new slope and the given point.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=-\\frac{3}{5}\\left(x-\\left(-4\\right)\\right)\\hfill \\\\ y - 1=-\\frac{3}{5}x-\\frac{12}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{12}{5}+\\frac{5}{5}\\hfill \\\\ y=-\\frac{3}{5}x-\\frac{7}{5}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\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-939\">\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>Revision and Adaptation. <strong>Provided 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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <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\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 1436. <strong>Authored by<\/strong>: WebWork-Rochester. <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>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 110960, 110970, 110971. <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>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/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":21,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at 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GPL\"}]","CANDELA_OUTCOMES_GUID":"b7a4605c-14de-428c-8df0-65168f2efe9e","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-939","chapter","type-chapter","status-publish","hentry"],"part":17,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/939","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":11,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/939\/revisions"}],"predecessor-version":[{"id":4018,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/939\/revisions\/4018"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/17"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/939\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=939"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=939"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=939"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=939"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}