{"id":6656,"date":"2020-10-03T16:30:11","date_gmt":"2020-10-03T16:30:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6656"},"modified":"2026-01-17T23:09:56","modified_gmt":"2026-01-17T23:09:56","slug":"3-7-writing-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/3-7-writing-linear-equations\/","title":{"raw":"3.7: Writing Linear Equations","rendered":"3.7: Writing Linear Equations"},"content":{"raw":"
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section 3.7 Learning Objectives<\/h3>\r\n3.7: Writing Linear Equations<\/strong>\r\n
    \r\n \t
  • Write the equation of the line given the slope and the y-intercept using the slope-intercept form<\/li>\r\n \t
  • Write the equation of the line given the slope and a point on the line using the point-slope form<\/li>\r\n \t
  • Write the equation of the line given two points on the line<\/li>\r\n \t
  • Write the equation of a horizontal or vertical line when given the graph<\/li>\r\n \t
  • Write the equation of a line given a point and a parallel or perpendicular line<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nIn this section, we will write the equation of a line given some information about the line.\u00a0 We will use two forms of a line to help us do this.\r\n

    The Slope-Intercept Form:\u00a0 [latex]y = mx + b[\/latex]<\/p>\r\n

    The Point-Slope Form:\u00a0\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n\r\n

    Write the equation of a line given the slope and y-intercept<\/h2>\r\nIn a previous section, we were introduced to the Slope-Intercept Form of a line.\u00a0 We learned that if an equation is written in this form, it is easy to identify the slope of the line and its y-intercept.\u00a0 In this section, we are being asked to write the equation of the line.\u00a0 If we know the slope and the y-intercept of a line, we can use the Slope-Intercept Form to help us do that.\r\n

    [latex]y = mx + b[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,m\\,\\,\\,\\,=\\,\\,\\,\\text{slope}\\\\(x,y)=\\,\\,\\,\\text{a point on the line}\\\\\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,=\\,\\,\\,\\text{the y value of the y-intercept}\\end{array}[\/latex]<\/p>\r\n\r\n

    \r\n

    Example 1<\/h3>\r\nWrite the equation of the line that has a slope of [latex] \\displaystyle \\frac{1}{2}[\/latex] and a y<\/i>-intercept of [latex](0,-5)[\/latex].\r\n\r\n[reveal-answer q=\"624715\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624715\"]Substitute the slope (m<\/i>) into [latex]y=mx+b[\/latex].\r\n

    [latex] \\displaystyle y=\\frac{1}{2}x+b[\/latex]<\/p>\r\nSubstitute the y<\/i>-intercept (b<\/i>) into the equation.\r\n

    [latex] \\displaystyle y=\\frac{1}{2}x-5[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]y=\\frac{1}{2}x-5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe can also find the equation by looking at a graph and finding the slope and y<\/em>-intercept.\r\n
    \r\n

    Example 2<\/h3>\r\nWrite the equation of the line in the graph by identifying the slope and y<\/em>-intercept.\r\n\"A\r\n[reveal-answer q=\"96446\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"96446\"]Identify the point where the graph crosses the y-axis [latex](0,3)[\/latex]. This means the y<\/em>-intercept is (0,3).\r\n\r\nIdentify one other point and draw a slope triangle to find the slope.\r\n\r\nThe slope is [latex]\\frac{-2}{3}[\/latex]\r\n\r\nSubstitute the slope and y<\/em> value of the intercept into the slope-intercept equation.\r\n

    [latex]y=mx+b\\\\y=\\frac{-2}{3}x+b\\\\y=\\frac{-2}{3}x+3[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]y=\\frac{-2}{3}x+3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nhttps:\/\/youtu.be\/GIn7vbB5AYo\r\n\r\n \r\n

    Write the equation of a line given the slope and a point on the line<\/h2>\r\nUsing the slope-intercept form of a line is easy when you know both the slope (m<\/i>) and the y<\/i>-intercept (b<\/i>), but what if you know the slope and just any point on the line, not specifically the y<\/i>-intercept? Can you still write the equation? The answer is yes<\/i>, but you will need to put in a little more thought and work than you did previously.\r\n

    Point-Slope Form<\/h3>\r\n

    [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,m\\,\\,\\,\\,=\\,\\,\\,\\text{slope}\\\\\\left({x}_{1},{y}_{1}\\right)=\\,\\,\\,\\text{a point on the line}\\end{array}[\/latex]<\/p>\r\nThis is an important formula, as it will be used in other algebra courses and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.\r\n

    \r\n

    Point-Slope Form<\/h3>\r\nGiven a point [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and slope m<\/em>, point-slope form will give the following equation of a line:\r\n

    [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn our first example, we are given the slope and a point on a line.\u00a0 We will use the Point-Slope Form to write the equation of the line.\r\n

    \r\n

    Example 3<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] that passes through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"524449\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"524449\"]\r\n\r\nUsing point-slope form, substitute [latex]-3[\/latex] for m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n

    [latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/p>\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows how to write the equation for a line given its slope and a point on the line.\r\n\r\nhttps:\/\/youtu.be\/vut5b2fRQQ0\r\n

    \r\n

    Think about it<\/h3>\r\nThere is an alternate method to finding the equation of a line given a point and a slope. See the example below.\r\n\r\nWrite the equation of the line that has a slope of 3 and contains the point [latex](1,4)[\/latex].<\/strong>\r\n\r\n[reveal-answer q=\"161353\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161353\"]\r\n\r\nRecall that a point is an (x<\/i>, y<\/i>) coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don\u2019t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[\/latex].\r\n\r\nYou do know the slope (m<\/i>), but you just don\u2019t know the value of the y<\/i>-intercept (b<\/i>). Since point (x<\/i>, y<\/i>) is a solution to the equation, you can substitute its coordinates for x<\/i> and y<\/i> in [latex]y=mx+b[\/latex]\u00a0and solve to find b<\/i>!\r\n\r\n \r\n\r\nSubstitute the slope (m<\/i>) into\u00a0[latex]y=mx+b[\/latex].\r\n

    [latex]y=3x+b[\/latex]<\/p>\r\nSubstitute the point [latex](1,4)[\/latex] for x <\/i>and y.<\/i>\r\n

    [latex]4=3\\left(1\\right)+b[\/latex]<\/p>\r\nSolve for b.<\/i>\r\n

    [latex]\\begin{array}{l}4=3+b\\\\1=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=3[\/latex]\u00a0and [latex]b=1[\/latex].\r\n

    Answer<\/h4>\r\n[latex]y=3x+1[\/latex]\r\n\r\nTo confirm our algebra, you can check by graphing the equation [latex]y=3x+1[\/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[\/latex].\r\n\r\n\"An\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video gives an example of this alternate method.\r\n\r\nhttps:\/\/youtu.be\/URYnKqEctgc\r\n

    Write the equation of a line given two points on the line<\/h2>\r\nLet\u2019s suppose you don\u2019t know either the slope or the y<\/i>-intercept, but you do know the location of two points on the line. It is more challenging, but you can find the equation of the line that would pass through those two points.\u00a0 We will begin by using the two points to find the slope of the line.\u00a0 Once you know the slope of the line, you can use the slope and one of the points in the Point-Slope Form to write the equation of the line.\r\n
    \r\n

    Example 4<\/h3>\r\nFind the equation of the line that passes through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"249539\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249539\"]\r\n\r\nFirst, we calculate the slope using the slope formula and two points.\r\n

    [latex]\\begin{array}{l}m\\hfill & =\\dfrac{-3 - 4}{0 - 3}\\hfill \\\\ \\hfill & =\\dfrac{-7}{-3}\\hfill \\\\ \\hfill & =\\dfrac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nThe easiest way to solve this problem is to recognize that the second point given, [latex](0,-3)[\/latex], is in fact the [latex]y[\/latex]-intercept. This means we know that [latex]b=-3[\/latex] in the slope-intercept form, [latex]y=mx+b[\/latex]. Plugging this value in along with the slope gives the desired equation.\r\n

    [latex]y=mx+b[\/latex]<\/p>\r\n

    [latex]y=\\dfrac{7}{3}x-3[\/latex]<\/p>\r\n\r\n

    Exploring Further<\/span><\/h4>\r\nAlthough we have now answered the question, we now use point-slope form to verify that this will produce the same answer. With the slope of [latex]\\dfrac{7}{3}[\/latex], we can use either point. Let us pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n

    [latex]\\begin{array}{l}y - 4=\\dfrac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\dfrac{7}{3}x - 7\\hfill&\\text{Distribute the }\\dfrac{7}{3}.\\hfill \\\\ y=\\dfrac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\r\nIn slope-intercept form, the equation is written as [latex]y=\\dfrac{7}{3}x - 3[\/latex].\r\n\r\nThis matches the result we arrived at earlier.\r\n

    Answer<\/span><\/h4>\r\n[latex]y=\\dfrac{7}{3}x-3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    In the next example, we again look for the equation of a line given two points. However, this time we are not given the [latex]y[\/latex]-intercept.<\/p>\r\n\r\n

    \r\n

    Example 5<\/h3>\r\nWrite the equation of the line that passes through the points [latex](2,1)[\/latex] and [latex](\u22121,\u22125)[\/latex]. Write the equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"977598\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"977598\"]\r\n\r\nFirst, find the slope using the given points.\r\n

    [latex] \\displaystyle \\frac{1-(-5)}{2-(-1)}=\\frac{6}{3}=2[\/latex]<\/p>\r\nNext, we plug this into point-slope form along with either point. Let us use the point [latex](2,1)[\/latex] for [latex](x_1,y_1)[\/latex]. Then, we solve for [latex]y[\/latex].\r\n

    [latex]\\begin{array}{l}y-1=2(x-2)\\\\y-1=2x-4\\\\y=2x-3\\end{array}[\/latex]<\/p>\r\n\r\n

    Exploring Further<\/span><\/h4>\r\nLet us take this opportunity to confirm our claim that we can use either point in the point-slope form. With the slope we found earlier, this time we use the point [latex](-1,-5)[\/latex] for [latex](x_1,y_1)[\/latex].\r\n

    [latex]\\begin{array}{l}y-(-5)=2(x-(-1))\\\\y+5=2(x+1)\\\\y+5=2x+2\\\\y=2x-3\\end{array}[\/latex]<\/p>\r\nBy converting to slope-intercept form, this verifies that either point (and in fact, any point on the line) will produce the same result.\r\n

    Answer<\/span><\/h4>\r\n[latex]y=2x-3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo further support that all the ideas we have presented will yield the same result, we now redo the same problem from Example 5 using the \"alternate method\" presented earlier in this section.\r\n
    \r\n

    Example 6<\/h3>\r\nWrite the equation of the line that passes through the points [latex](2,1)[\/latex] and [latex](\u22121,\u22125)[\/latex].\u00a0Write the equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"333536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333536\"]\r\n\r\nFirst, find the slope using the given points.\r\n

    [latex] \\displaystyle \\frac{1-(-5)}{2-(-1)}=\\frac{6}{3}=2[\/latex]<\/p>\r\nSubstitute the slope, [latex]m[\/latex], into [latex]y=mx+b[\/latex].\r\n

    [latex]y=2x+b[\/latex]<\/p>\r\nSubstitute the coordinates of either point for [latex]x[\/latex]\u00a0<\/i>and\u00a0[latex]y[\/latex] \u2013 this example uses [latex](2, 1)[\/latex].\r\n

    [latex]1=2(2)+b[\/latex]<\/p>\r\nSolve for\u00a0[latex]b[\/latex].\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,1=4+b\\\\\u22123=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=2[\/latex] and [latex]b=-3[\/latex].\r\n

    [latex]y=2x-3[\/latex]<\/p>\r\nNote that this matches the answer we obtained using point-slope form in Example 5.\r\n

    Answer<\/h4>\r\n[latex]y=2x-3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that is doesn\u2019t matter which point you use when you substitute and solve for b<\/i>\u2014you get the same result for b<\/i> either way. In the example above, you substituted the coordinates of the point (2, 1) in the equation [latex]y=2x+b[\/latex]. Let\u2019s start with the same equation, [latex]y=2x+b[\/latex], but substitute in [latex](\u22121,\u22125)[\/latex]:\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,y=2x+b\\\\-5=2\\left(-1\\right)+b\\\\-5=-2+b\\\\-3=b\\end{array}[\/latex]<\/p>\r\nThe final equation is the same: [latex]y=2x\u20133[\/latex].\r\n

    Video: Write the equation of a line given two points on the line (Alternate Method)<\/h3>\r\nhttps:\/\/youtu.be\/P1ex_a6iYDo\r\n\r\n \r\n

    Write the equation of a line given a point and a parallel or perpendicular line<\/h2>\r\nThe relationships between slopes of parallel and perpendicular lines can be used to write equations of parallel and perpendicular lines.\r\n\r\nLet\u2019s start with an example involving parallel lines.\r\n

    Write the equation of a line given a point and a parallel line<\/h3>\r\n
    \r\n

    Example 7<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]x\u2013y=5[\/latex] and goes through the point [latex](\u22122,1)[\/latex].\r\n\r\n[reveal-answer q=\"763534\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"763534\"]\r\n\r\nRewrite the line you want to be parallel to into the\u00a0[latex]y=mx+b[\/latex] form, if needed.\r\n

    [latex]\\begin{array}{r}x\u2013y=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\u2212y=\u2212x+5\\\\y=x\u20135\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=1[\/latex].\u00a0 Therefore, the slope is 1.\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is 1.\r\n\r\nUse the method for writing an equation from the slope and a point on the line. We will use the Point-Slope Form and substitute 1 for m<\/i>, and the point [latex](\u22122,1)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] .\r\n\r\n \r\n

    [latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\y - 1=1\\left(x - (-2)\\right)\\hfill \\\\y - 1=1\\left(x + 2\\right)\\hfill \\\\ y - 1=x + 2 \\\\ y=x + 3\\hfill \\end{array}[\/latex]<\/span><\/p>\r\n \r\n\r\nAnswer<\/span>\r\n\r\n[latex]y=x+3[\/latex][\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n

    <\/h2>\r\nhttps:\/\/youtu.be\/TQKz2XHI09E\r\n

    Write the equation of a line given a point and a perpendicular line<\/h3>\r\nWhen you are working with perpendicular lines, you will usually be given one of the lines and an additional point. Remember that two non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other.\u00a0To find the slope of a perpendicular line, find the reciprocal, and then find the opposite of this reciprocal. \u00a0In other words, flip it and change the sign.\r\n
    \r\n

    Example 8<\/h3>\r\nWrite the equation of a line that contains the point [latex](1,5)[\/latex] and is perpendicular to the line [latex]y=2x\u2013 6[\/latex].\r\n\r\n[reveal-answer q=\"604282\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"604282\"]\r\n\r\nIdentify the slope of the line you want to be perpendicular to.\r\n\r\nThe given line is written in [latex]y=mx+b[\/latex] form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope is 2.\r\n\r\nTo find the slope of a perpendicular line, find the reciprocal, [latex] \\displaystyle \\frac{1}{2}[\/latex], then the opposite, [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nThe slope of the perpendicular line is [latex] \\displaystyle -\\frac{1}{2}[\/latex].\r\n\r\nFor practice, this time, we will use the alternate method for finding the equation of a line. Substitute [latex] \\displaystyle -\\frac{1}{2}[\/latex] for [latex]m[\/latex], and the point [latex](1,5)[\/latex] for [latex]x[\/latex] and [latex]y[\/latex] into slope-intercept form.\r\n

    [latex] \\displaystyle \\begin{array}{l}y=mx+b\\\\5=-\\frac{1}{2}(1)+b\\end{array}[\/latex]<\/p>\r\nSolve for [latex]b[\/latex].\r\n

    [latex] \\displaystyle \\begin{array}{l}\\,\\,\\,5=-\\frac{1}{2}+b\\\\\\frac{11}{2}=b\\end{array}[\/latex]<\/p>\r\nWrite the equation using the new slope for [latex]m[\/latex] and the [latex]b[\/latex] you just found.\r\n

    [latex]y=mx+b[\/latex]<\/p>\r\n

    [latex]y=-\\dfrac{1}{2}x+\\frac{11}{2}[\/latex]<\/p>\r\nNote, we could have used point-slope form to produce this same result as well. You are encouraged to verify this.\r\n

    Answer<\/span><\/h4>\r\n[latex]y=-\\frac{1}{2}x+\\frac{11}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n

    Video: Write the equation of a line given a point and a perpendicular line (Alternate Method)<\/h3>\r\nhttps:\/\/youtu.be\/QtvtzKjtowA\r\n

    Write the equations of lines parallel and perpendicular to horizontal and vertical lines<\/span><\/h2>\r\n
    \r\n

    Example 9<\/h3>\r\nWrite the equation of a line that is parallel to the line [latex]y=4[\/latex] through the point [latex](0,10)[\/latex].\r\n\r\n[reveal-answer q=\"426450\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426450\"]\r\n\r\nRewrite the line into [latex]y=mx+b[\/latex]\u00a0form, if needed.\r\n\r\nYou may notice without doing this that [latex]y=4[\/latex]\u00a0is a horizontal line 4 units above the x<\/i>-axis. Because it is horizontal, you know its slope is zero.\r\n

    [latex]\\begin{array}{l}y=4\\\\y=0x+4\\end{array}[\/latex]<\/p>\r\nIdentify the slope of the given line.\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=4[\/latex].\r\n\r\nSince [latex]m=0[\/latex], the slope is 0. This is a horizontal line.\r\n\r\nTo find the slope of a parallel line, use the same slope.\r\n\r\nThe slope of the parallel line is also 0.\r\n\r\nSince the parallel line will be a horizontal line, its form is\r\n

    [latex]y=\\text{a constant}[\/latex]<\/p>\r\nSince we want this new line to pass through the point [latex](0,10)[\/latex], we will need to write the equation of the new line as:\r\n

    [latex]y=10[\/latex]<\/p>\r\nThis line is parallel to [latex]y=4[\/latex]\u00a0and passes through [latex](0,10)[\/latex].\r\n

    Answer<\/h4>\r\n[latex]y=10[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example 10<\/h3>\r\nWrite the equation of a line that is perpendicular to the line [latex]y=-3[\/latex] through the point [latex](-2,5)[\/latex].\r\n\r\n[reveal-answer q=\"426550\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426550\"]\r\n\r\nIn the equation above, [latex]m=0[\/latex] and [latex]b=-3[\/latex].\r\n\r\nA perpendicular line will have a slope that is the negative reciprocal of the slope of\u00a0[latex]y=-3[\/latex], but\u00a0what does that mean in this case?\r\n\r\nThe reciprocal of 0 is [latex]\\frac{1}{0}[\/latex], but we know that dividing by 0 is undefined.\r\n\r\nThis means that we are looking for a line whose slope is undefined, and we also know that vertical lines have slopes that are undefined. This makes sense since we started with a horizontal line.\r\n\r\nThe form of a vertical line is [latex]x=\\text{a constant}[\/latex], where every x<\/em>-value on the line is equal to some constant. \u00a0Since we are looking for a line that goes through the point [latex](-2,5)[\/latex], all of the x<\/em>-values on this line must be [latex]-2[\/latex].\r\n\r\nThe equation of a line passing through [latex](-2,5)[\/latex] that is perpendicular to the horizontal line\u00a0[latex]y=-3[\/latex] is therefore,\r\n

    [latex]x=-2[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=-2[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n

    <\/h2>\r\nhttps:\/\/youtu.be\/Qpn3f3wMeIs\r\n

    Summary<\/span><\/h2>\r\nThe slope-intercept form of a linear equation is written as [latex]y=mx+b[\/latex], where m<\/i> is the slope and b<\/i> is the value of y<\/i> at the y<\/i>-intercept, which can be written as [latex](0,b)[\/latex]. When you know the slope and the y<\/i>-intercept of a line you can use the slope-intercept form to immediately write the equation of that line. The point-slope form,\u00a0\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex], can be used to write the equation of a line when you know the slope and a point on the line or when you know two points on the line.\r\n\r\nWhen lines in a plane are parallel (that is, they never cross), they have the same slope. When lines are perpendicular (that is, they cross at a 90\u00b0 angle), their slopes are opposite reciprocals of each other. The product of their slopes will be [latex]-1[\/latex], except in the case where one of the lines is vertical causing its slope to be undefined. You can use these relationships to find an equation of a line that goes through a particular point and is parallel or perpendicular to another line.\r\n

    <\/h2>","rendered":"
    \n

    section 3.7 Learning Objectives<\/h3>\n

    3.7: Writing Linear Equations<\/strong><\/p>\n

      \n
    • Write the equation of the line given the slope and the y-intercept using the slope-intercept form<\/li>\n
    • Write the equation of the line given the slope and a point on the line using the point-slope form<\/li>\n
    • Write the equation of the line given two points on the line<\/li>\n
    • Write the equation of a horizontal or vertical line when given the graph<\/li>\n
    • Write the equation of a line given a point and a parallel or perpendicular line<\/li>\n<\/ul>\n<\/div>\n

       <\/p>\n

      In this section, we will write the equation of a line given some information about the line.\u00a0 We will use two forms of a line to help us do this.<\/p>\n

      The Slope-Intercept Form:\u00a0 [latex]y = mx + b[\/latex]<\/p>\n

      The Point-Slope Form:\u00a0\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n

      Write the equation of a line given the slope and y-intercept<\/h2>\n

      In a previous section, we were introduced to the Slope-Intercept Form of a line.\u00a0 We learned that if an equation is written in this form, it is easy to identify the slope of the line and its y-intercept.\u00a0 In this section, we are being asked to write the equation of the line.\u00a0 If we know the slope and the y-intercept of a line, we can use the Slope-Intercept Form to help us do that.<\/p>\n

      [latex]y = mx + b[\/latex]<\/p>\n

      [latex]\\begin{array}{l}\\,\\,\\,\\,\\,m\\,\\,\\,\\,=\\,\\,\\,\\text{slope}\\\\(x,y)=\\,\\,\\,\\text{a point on the line}\\\\\\,\\,\\,\\,\\,\\,\\,b\\,\\,\\,\\,=\\,\\,\\,\\text{the y value of the y-intercept}\\end{array}[\/latex]<\/p>\n

      \n

      Example 1<\/h3>\n

      Write the equation of the line that has a slope of [latex]\\displaystyle \\frac{1}{2}[\/latex] and a y<\/i>-intercept of [latex](0,-5)[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      Substitute the slope (m<\/i>) into [latex]y=mx+b[\/latex].<\/p>\n

      [latex]\\displaystyle y=\\frac{1}{2}x+b[\/latex]<\/p>\n

      Substitute the y<\/i>-intercept (b<\/i>) into the equation.<\/p>\n

      [latex]\\displaystyle y=\\frac{1}{2}x-5[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]y=\\frac{1}{2}x-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      We can also find the equation by looking at a graph and finding the slope and y<\/em>-intercept.<\/p>\n

      \n

      Example 2<\/h3>\n

      Write the equation of the line in the graph by identifying the slope and y<\/em>-intercept.
      \n\"A<\/p>\n

      Show Solution<\/span><\/p>\n
      Identify the point where the graph crosses the y-axis [latex](0,3)[\/latex]. This means the y<\/em>-intercept is (0,3).<\/p>\n

      Identify one other point and draw a slope triangle to find the slope.<\/p>\n

      The slope is [latex]\\frac{-2}{3}[\/latex]<\/p>\n

      Substitute the slope and y<\/em> value of the intercept into the slope-intercept equation.<\/p>\n

      [latex]y=mx+b\\\\y=\\frac{-2}{3}x+b\\\\y=\\frac{-2}{3}x+3[\/latex]<\/p>\n

      Answer<\/h4>\n

      [latex]y=\\frac{-2}{3}x+3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n