{"id":6648,"date":"2020-10-03T15:58:15","date_gmt":"2020-10-03T15:58:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6648"},"modified":"2026-01-17T06:15:02","modified_gmt":"2026-01-17T06:15:02","slug":"3-4-slope-of-a-line","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/3-4-slope-of-a-line\/","title":{"raw":"3.4: Slope of a Line","rendered":"3.4: Slope of a Line"},"content":{"raw":"
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section 3.4 Learning Objectives<\/h3>\r\n3.4: Slope of a Line<\/strong>\r\n
    \r\n \t
  • Identify the slope of a line from the graph<\/li>\r\n \t
  • Find the slope of a line given two points on a line<\/li>\r\n \t
  • Find the slope and y-intercept given the equation of a line<\/li>\r\n \t
  • Find the slope of horizontal and vertical lines<\/li>\r\n \t
  • Find the slopes of parallel and perpendicular lines<\/li>\r\n \t
  • Graph a line using the slope and y-intercept<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    Identify slope from a graph<\/h2>\r\nThe mathematical definition of slope<\/b> is very similar to our everyday one. In math, slope is used to describe the steepness and direction of lines. By just looking at the graph of a line, you can learn some things about its slope, especially relative to other lines graphed on the same coordinate plane. Consider the graphs of the three lines shown below:\r\n\r\n\"Coordinate\r\n\r\nFirst, let\u2019s look at lines A and B. If you imagined these lines to be hills, you would say that line B is steeper than line A. Line B has a greater slope than line A.\r\n\r\nNext, notice that lines A and B slant up as you move from left to right. We say these two lines have a positive slope. Line C slants down from left to right. Line C has a negative slope. Using two of the points on the line, you can find the slope of the line by finding the rise and the run. The vertical change between two points is called the rise<\/b>, and the horizontal change is called the run<\/b>. The slope equals the rise divided by the run: [latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].\r\n\r\n\"A\r\n\r\nYou can determine the slope of a line from its graph by looking at the rise and run. One characteristic of a line is that its slope is constant all the way along it. So, you can choose any 2 points along the graph of the line to figure out the slope. Let\u2019s look at an example.\r\n
    \r\n

    Example 1<\/h3>\r\nUse the graph to find the slope of the line.\r\n\r\n\"A<\/b>\r\n\r\n[reveal-answer q=\"606472\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"606472\"]\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\text{rise}=2[\/latex]<\/td>\r\nStart from a point on the line, such as [latex](2,1)[\/latex] and move vertically until in line with another point on the line, such as [latex](6,3)[\/latex]. The rise is 2 units. It is positive because you moved up.<\/td>\r\n<\/tr>\r\n
    [latex]\\text{run}=4[\/latex]<\/td>\r\nNext, move horizontally to the point [latex](6,3)[\/latex]. Count the number of units. The run is 4 units. It is positive because you moved to the right.<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\text{Slope}=\\frac{2}{4}=\\frac{1}{2}[\/latex]<\/td>\r\n[latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\n[latex]Slope=\\frac{1}{2}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThis line will have a slope of [latex] \\displaystyle \\frac{1}{2}[\/latex] no matter which two points you pick on the line. Try measuring the slope from the origin, [latex](0,0)[\/latex], to the point [latex](6,3)[\/latex]. You will find that the [latex]\\text{rise}=3[\/latex] and the [latex]\\text{run}=6[\/latex]. The slope is [latex] \\displaystyle \\frac{\\text{rise}}{\\text{run}}=\\frac{3}{6}=\\frac{1}{2}[\/latex]. It is the same!\r\n\r\nLet\u2019s look at another example.\r\n
    \r\n

    Example 2<\/h3>\r\nUse the graph to find the slope of the two lines.\u00a0<\/b>\r\n\r\n\"Two<\/b>\r\n\r\n[reveal-answer q=\"860733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"860733\"]Notice that both of these lines have positive slopes, so you expect your answers to be positive.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Blue line<\/th>\r\n<\/tr>\r\n
    [latex]\\text{rise}=4[\/latex]<\/td>\r\nStart with the blue line, going from point [latex](-2,1)[\/latex] to point [latex](-1,5)[\/latex]. This line has a rise of 4 units up, so it is positive.<\/td>\r\n<\/tr>\r\n
    [latex]\\text{run}=1[\/latex]<\/td>\r\nRun is 1 unit to the right, so it is positive.<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\text{Slope }=\\frac{4}{1}=4[\/latex]<\/td>\r\nSubstitute the values for the rise and run in the formula [latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/td>\r\n<\/tr>\r\n
    Red line<\/th>\r\n<\/tr>\r\n
    [latex]\\text{rise}=1[\/latex]<\/td>\r\nThe red line, going from point [latex](-1,-2)[\/latex] to point [latex](3,-1)[\/latex] has a rise of 1 unit.<\/td>\r\n<\/tr>\r\n
    [latex]\\text{run}=4[\/latex]<\/td>\r\nThe red line has a run of 4 units.<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\text{Slope }=\\frac{1}{4}[\/latex]<\/td>\r\nSubstitute the values for the rise and run into the formula [latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope of the blue line is 4 and the slope of the red line is [latex]\\frac{1}{4}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nWhen you look at the two lines, you can see that the blue line is steeper than the red line. It makes sense the value of the slope of the blue line, 4, is greater than the value of the slope of the red line, [latex] \\displaystyle \\frac{1}{4}[\/latex]. The greater the slope, the steeper the line.\r\n\r\nhttps:\/\/youtu.be\/29BpBqsiE5w\r\n

    Distinguish between graphs of lines with negative and positive slopes<\/h3>\r\nDirection is important when it comes to determining slope. It\u2019s important to pay attention to whether you are moving up, down, left, or right; that is, if you are moving in a positive or negative direction. If you go up to get to your second point, the rise is positive. If you go down to get to your second point, the rise is negative. If you go right to get to your second point, the run is positive. If you go left to get to your second point, the run is negative.\r\n\r\nIn the following two examples, you will see a slope that is positive and one that is negative.\r\n
    \r\n

    Example 3 (Advanced)<\/h3>\r\nFind the slope of the line graphed below.\r\n\r\n\"Line<\/b>\r\n\r\n[reveal-answer q=\"82644\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"82644\"]\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\text{rise}=4.5[\/latex]<\/td>\r\nStart at [latex](-3,-0.25)[\/latex] and rise 4.5. This means moving 4.5 units in a positive direction.<\/td>\r\n<\/tr>\r\n
    [latex]\\text{run}=6[\/latex]<\/td>\r\nFrom there, run 6 units in a positive direction to [latex](3,4.25)[\/latex].<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\text{Slope}=\\frac{4.5}{6}=0.75[\/latex]<\/td>\r\n[latex]\\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope of the line is 0.75.[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next example shows a line with a negative slope.\r\n
    \r\n

    Example 4<\/h3>\r\nFind the slope of the line graphed below.\r\n\r\n\"A\r\n\r\n[reveal-answer q=\"924393\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924393\"]\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\text{rise}=-3[\/latex]<\/td>\r\nStarting at Point A, [latex](0,4)[\/latex], we must go down 3 units to line up with Point B. This means the rise is [latex]-3[\/latex] since we move 3 units in a negative direction.<\/td>\r\n<\/tr>\r\n
    [latex]\\text{run}=2[\/latex]<\/td>\r\nFrom there, run 2 units in a positive direction to Point B [latex](2,1)[\/latex].<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\text{Slope}=\\frac{-3}{2}[\/latex]<\/td>\r\n[latex] \\displaystyle \\text{Slope}=\\frac{\\text{rise}}{\\text{run}}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope of the line is [latex]-\\frac{3}{2}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the example above, you could have also found the slope by starting at point B, rising [latex]+3[\/latex] (up) and then running [latex]-2[\/latex] (left) to arrive at point A. The result is still a slope of [latex]\\displaystyle\\frac{\\text{rise}}{\\text{run}}=\\frac{+3}{-2}=-\\frac{3}{2}[\/latex].\r\n

    Find the slope given two points on the line<\/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 line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an x<\/i>-value and a y<\/i>-value, written as an ordered pair (x<\/i>, y<\/i>). The x<\/i>-value tells you where a point is horizontally. The y-<\/i>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\"A\r\n\r\nThe rise is the vertical change between the two points, which is the difference between their y<\/i>-coordinates. That makes the rise [latex]\\left(y_{2}-y_{1}\\right)[\/latex]. The run between these two points is the difference in the x<\/i>-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\nIn the example below, you\u2019ll see that the line has two points each indicated as an ordered pair. 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.\r\n\r\n\"Coordinate\r\n\r\nYou can see from the graph that the rise going from Point 1 to Point 2 is 4, because you are moving 4 units in a positive direction (up). The run is [latex]\u22122[\/latex], because you are then moving in a negative direction (left) 2 units. Using the slope formula,\r\n

    [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\r\n\r\n\r\n\r\n\r\n\r\n
    Name<\/th>\r\nOrdered Pair<\/th>\r\nCoordinates<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    Point 1<\/td>\r\n[latex](0,2)[\/latex]<\/td>\r\n[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n
    Point 2<\/td>\r\n[latex](\u22122,6)[\/latex]<\/td>\r\n[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, m<\/i>, is [latex]\u22122[\/latex].\r\n\r\nIt 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.\r\n
    \r\n

    Example 5<\/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=\"666697\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666697\"]\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]\\begin{array}{l}x_{1}=4\\\\y_{1}=2\\end{array}[\/latex]<\/td>\r\n[latex]\\left(4,2\\right)=\\text{Point }1,\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\begin{array}{l}x_{2}=5\\\\y_{2}=5\\end{array}[\/latex]<\/td>\r\n[latex]\\left(5,5\\right)=\\text{Point }2,\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
    [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\nSubstitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope is 3.[\/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
    \r\n

    Example 6<\/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\r\n\r\n\r\n\r\n\r\n
    [latex]\\begin{array}{l}x_{1}=5\\\\y_{1}=5\\end{array}[\/latex]<\/td>\r\n[latex](5,5)=\\text{Point }1[\/latex], [latex]\\left(x_{1},y_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=2\\end{array}[\/latex]<\/td>\r\n[latex](4,2)=\\text{Point }2[\/latex], [latex]\\left(x_{2},y_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
    [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\nSubstitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope is 3.[\/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 3.\r\n
    \r\n

    Example 7 (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\r\n\r\n\r\n\r\n\r\n
    [latex]\\begin{array}{l}x_{1}=3\\\\y_{1}=-6.25\\end{array}[\/latex]<\/td>\r\n[latex](3,-6.25)=\\text{Point }1[\/latex], [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-1\\\\{{y}_{2}}=8.5\\end{array}[\/latex]<\/td>\r\n[latex](-1,8.5)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n
    [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\nSubstitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope is [latex]-3.6875[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n

    <\/h2>\r\nhttps:\/\/youtu.be\/ZW7rQa8SJSU\r\n

    Find the slope and y-intercept given the equation of the line:<\/strong><\/h2>\r\nWe can also determine the slope of a line from the equation of the line.\u00a0 To do this, we first need to write the equation in Slope-Intercept Form.\r\n

    Slope-Intercept Form: [latex]y=mx+b[\/latex]\u00a0 \u00a0<\/strong><\/p>\r\nTo write an equation in Slope-Intercept Form, we need to solve for [latex]y[\/latex] in the equation.\u00a0 Once the equation is in this form, we know a couple of things just by looking at the equation.\r\n

    The coefficient of [latex]x[\/latex] is the slope of the line.<\/strong><\/p>\r\n

    The y-intercept is the point [latex](0,b)[\/latex].<\/strong><\/p>\r\n\r\n

    \r\n

    Example 8<\/h3>\r\nFind the slope and y-intercept of the line with equation [latex]y=-3x+7[\/latex]\r\n\r\n[reveal-answer q=\"16868\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"16868\"]\r\n\r\nThis equation is already in Slope-Intercept Form.\u00a0 Therefore, to find the slope of the line we just need to identify the coefficient of [latex]x[\/latex].\u00a0 In this equation, that is [latex]-3[\/latex].\r\n

    [latex]y=-3x+7[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 Slope =\u00a0 [latex]-3[\/latex]<\/p>\r\nThe y-intercept of a line is always a point with an x-coordinate of 0.\u00a0 We can find the y-coordinate of the y-intercept by looking at the constant term in the equation when it is in Slope-Intercept Form.\u00a0 In this equation, that is 7. The reason this works is because if we were to substitute [latex]x=0[\/latex] into the equation, we would quickly obtain an output of [latex]y=7[\/latex].\r\n

    [latex]y=-3x+7[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0y-intercept =\u00a0 [latex](0, 7)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Example 9<\/h3>\r\nFind the slope and y-intercept of the line with equation\u00a0[latex]-4x+2y=-6[\/latex]\r\n\r\n[reveal-answer q=\"177985\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"177985\"]\r\n\r\nWe first need to solve for [latex]y[\/latex] to write this equation in Slope-Intercept Form.\r\n

    [latex]-4x+2y=-6[\/latex]<\/p>\r\nAdd\u00a0[latex]4x[\/latex] to both sides of the equation to get:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]2y=4x-6[\/latex]\r\n\r\n \r\n\r\nDivide each term by 2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex] \\frac{2y}{2}=\\frac{4x}{2}-\\frac{6}{2} [\/latex]\r\n

    [latex]y=2x-3[\/latex]<\/p>\r\nNow that the equation is in Slope-Intercept Form, we can see that the slope is 2 and the y-intercept is the point [latex](0,-3)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    \r\n

    Example 10<\/h3>\r\nFind the slope and y-intercept of the line with equation [latex]y=\\frac{3}{5}x[\/latex]:\r\n\r\n[reveal-answer q=\"121959\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"121959\"]\r\n\r\nThis equation is already in Slope-Intercept Form.\u00a0 Therefore, the slope is\u00a0[latex]\\frac{3}{5}[\/latex].\u00a0 However, what would the y-intercept be?\r\n\r\nNote that the equation can be rewritten as [latex]y=\\frac{3}{5}x+0[\/latex]. Written this way it is easier to see that the y-intercept would be the point [latex](0,0)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Find the slope of horizontal and vertical lines<\/h2>\r\nSo far you\u2019ve considered lines that run \u201cuphill\u201d or \u201cdownhill.\u201d Their slopes may be steep or gradual, but they are always positive or negative numbers. But there are two other kinds of lines, horizontal and vertical. What is the slope of a flat line or level ground? Of a wall or a vertical line?\r\n\r\n\"Horizontal\r\n\r\nLet\u2019s consider a horizontal line on a graph. No matter which two points you choose on the line, they will always have the same y<\/i>-coordinate. The equation for this line is [latex]y=3[\/latex]. The equation can also be written as [latex]y=\\left(0\\right)x+3[\/latex].\r\n\r\nUsing the form [latex]y=0x+3[\/latex], you can see that the slope is 0. 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

    [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 0.\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 y<\/i>-coordinate. So, when you apply the slope formula, the numerator will always be 0. Zero divided by any non-zero number is 0, so the slope of any horizontal line is always 0.\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 y-<\/i>coordinate will always be 3.\r\n\r\nHow about vertical lines? In this case, no matter which two points you choose, they will always have the same x<\/i>-coordinate. The equation for the line shown below is [latex]x=2[\/latex].\r\n\r\n\"Vertical\r\n\r\nThere is no way that this equation can be put in the slope-point form, as the coefficient of y<\/i> is [latex]0[\/latex].\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

    [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 is undefined 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

    \r\n

    Example 11<\/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\r\n\r\n\r\n\r\n\r\n
    [latex] \\displaystyle \\begin{array}{l}{{x}_{1}}=3\\\\{{y}_{1}}=2\\end{array}[\/latex]<\/td>\r\n[latex](3,2)=\\text{Point }1[\/latex], [latex] \\displaystyle \\left(x_{1},x_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-8\\\\\\\\{{y}_{2}}=2\\end{array}[\/latex]<\/td>\r\n[latex](\u22128,2)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n
    [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\nSubstitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    Answer<\/h4>\r\nThe slope is 0, so the line is horizontal.[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/youtu.be\/zoLM3rxzndo\r\n
    \r\n

    Example 12<\/h3>\r\nFind the slope and y-intercept of the line with equation\u00a0[latex]y=-2[\/latex] :\r\n\r\n[reveal-answer q=\"851789\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"851789\"]\r\n\r\nNote that the equation can be rewritten as [latex]y=0x-2[\/latex] :\r\n\r\nWritten this way it is easier to see that the slope is 0 and the y-intercept is\u00a0[latex](0, -2)[\/latex] .\r\n\r\n \r\n\r\nIf we were to graph [latex]y=-2[\/latex] , it would be a horizontal line.\u00a0 All horizontal lines have a slope of 0.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    ExAMPLE 13<\/h3>\r\nFind the slope and y-intercept of the line with equation [latex]x=3[\/latex]:\r\n\r\n[reveal-answer q=\"369359\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"369359\"]\r\n\r\nThis equation cannot be written in Slope-Intercept Form since there is no [latex]y[\/latex] term.\u00a0 To find the slope of the line, let\u2019s look at two points that are solutions to this equation.\u00a0 Any point with an x-coordinate of 3 is a solution to this equation.\u00a0 Therefore, both \u00a0[latex](3,0)[\/latex]\u00a0and\u00a0\u00a0[latex](3,-2)[\/latex] \u00a0are solutions to the equation [latex]x=3[\/latex].\r\n\r\n \r\n\r\nLet\u2019s use these two points to find the slope of the line.\r\n

    \u00a0[latex]m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{-2-1}{3-3}=\\frac{-2}{0}[\/latex]<\/p>\r\nNotice that we get 0 in the denominator.\u00a0 Since we cannot divide by 0, this slope is undefined<\/strong>.\r\n\r\nNext, let\u2019s consider the graph of [latex]x=3[\/latex].\r\n\r\n \r\n\r\nThe graph of\u00a0[latex]x=3[\/latex] is a vertical line.\u00a0 All vertical lines have a slope that is undefined.\r\n\r\n\"Graph\r\n\r\nWhat about the y-intercept?\u00a0 The y-intercept is the point where the line crosses the y-axis.\u00a0 Since the vertical line of [latex]x=3[\/latex]\u00a0does not cross the y-axis, there is no y-intercept.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Characterize the slopes of parallel and perpendicular lines<\/h2>\r\nWhen you graph two or more linear equations in a coordinate plane, they generally cross at a point. However, when two lines in a coordinate plane never cross, they are called parallel lines<\/b>. You will also look at the case where two lines in a coordinate plane cross at a right angle. These are called perpendicular lines<\/b>. The slopes of the graphs in each of these cases have a special relationship to each other.\r\n\r\nParallel lines are two or more lines in a plane that never intersect. Examples of parallel lines are all around us, such as the opposite sides of a rectangular picture frame and the shelves of a bookcase.\r\n\r\n\"Line\r\n\r\nPerpendicular lines are two or more lines that intersect at a 90-degree angle, like the two lines drawn on this graph. These 90-degree angles are also known as right angles.\r\n\r\n\"Two\r\n\r\nPerpendicular lines are also everywhere, not just on graph paper but also in the world around us, from the crossing pattern of roads at an intersection to the colored lines of a plaid shirt.\r\n
    \r\n

    Parallel Lines<\/h3>\r\nTwo non-vertical lines in a plane are parallel if they have both:\r\n
      \r\n \t
    • the same slope<\/li>\r\n \t
    • different y<\/i>-intercepts<\/li>\r\n<\/ul>\r\nAny two vertical lines in a plane are parallel.\r\n\r\n<\/div>\r\n

      Find the slopes of parallel lines<\/span><\/h2>\r\n
      \r\n

      Example 14<\/h3>\r\nFind the slope of a line parallel to the line [latex]y=\u22123x+4[\/latex].\r\n\r\n[reveal-answer q=\"350329\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350329\"]\r\n\r\nIdentify the slope of the given line.\r\n\r\nThe given line is written in [latex]y=mx+b[\/latex]\u00a0form, with [latex]m=\u22123[\/latex] and [latex]b=4[\/latex]. The slope is [latex]\u22123[\/latex].\r\n\r\nA line parallel to the given line has the same slope.\r\n

      Answer<\/h4>\r\nThe slope of the parallel line is [latex]\u22123[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n
      \r\n

      Example 15<\/h3>\r\nDetermine whether the lines [latex]y=6x+5[\/latex] and [latex]y=6x\u20131[\/latex]\u00a0are parallel.\r\n\r\n[reveal-answer q=\"619259\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"619259\"]\r\n\r\nIdentify the slopes of the given lines.\r\n\r\nThe given lines are written in [latex]y=mx+b[\/latex]\u00a0form, with [latex]m=6[\/latex]\u00a0for the first line and [latex]m=6[\/latex]\u00a0for the second line. The slope of both lines is 6.\r\n\r\nLook at b<\/i>, the y<\/i>-value of the y<\/i>-intercept, to see if the lines are perhaps exactly the same line, in which case we don\u2019t say they are parallel.\r\n\r\nThe first line has a y<\/i>-intercept at [latex](0,5)[\/latex], and the second line has a y<\/em>-intercept at [latex](0,\u22121)[\/latex]. They are not the same line.\r\n\r\nThe slopes of the lines are the same and they have different y<\/i>-intercepts, so they are not the same line and they are parallel.\r\n

      Answer<\/h4>\r\nThe lines are parallel.[\/hidden-answer]\r\n\r\n<\/div>\r\n

      Find the slopes of perpendicular lines<\/span><\/h2>\r\n
      \r\n

      Perpendicular Lines<\/h3>\r\nTwo non-vertical lines are perpendicular if the slope of one is the negative reciprocal of the slope of the other. If the slope of the first equation is [latex]\\frac{2}{3}[\/latex], then the slope of the second equation will need to be [latex]-\\frac{3}{2}[\/latex] for the lines to be perpendicular.\r\n\r\n<\/div>\r\nYou can also check the two slopes to see if the lines are perpendicular by multiplying the two slopes together. If they are perpendicular, the product of the slopes will be [latex]\u22121[\/latex]. For example, [latex] \\frac{2}{3}\\cdot-\\frac{3}{2}=-1[\/latex].\r\n
      \r\n

      Example 16<\/h3>\r\nFind the slope of a line perpendicular to the line [latex]y=2x\u20136[\/latex].\r\n\r\n[reveal-answer q=\"331107\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331107\"]The given line is written in\u00a0[latex]y=mx+b[\/latex]<\/span>\u00a0form, with [latex]m=2[\/latex] and [latex]b=-6[\/latex]. The slope of this line is then 2, or equivalently, [latex]m=\\frac{2}{1}[\/latex].\r\n\r\nA line perpendicular to this would have a slope that is the negative reciprocal of [latex]\\frac{2}{1}[\/latex], which is [latex]-\\frac{1}{2}[\/latex].\r\n

      Answer<\/h4>\r\nThe slope of the perpendicular line is [latex]-\\frac{1}{2}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nTo find the slope of a perpendicular line, find the reciprocal, [latex] \\displaystyle \\frac{2}{1}[\/latex], which is [latex]\\frac{1}{2}[\/latex],and then find the opposite of this reciprocal, which is [latex] \\displaystyle -\\tfrac{1}{2}[\/latex].\r\n\r\nNote that the product [latex]2\\left(-\\frac{1}{2}\\right)=\\frac{2}{1}\\left(-\\frac{1}{2}\\right)=-1[\/latex], so this means the lines are perpendicular.\r\n\r\nIn the case where one of the lines is vertical, the slope of that line is undefined and it is not possible to calculate the product with an undefined number. When one line is vertical, the line perpendicular to it will be horizontal, having a slope of zero ([latex]m=0[\/latex]).\r\n
      \r\n

      Example 17<\/h3>\r\nDetermine whether the lines [latex]y=\u22128x+5[\/latex]\u00a0and [latex] \\displaystyle y\\,\\text{=}\\,\\,\\frac{1}{8}x-1[\/latex] are parallel, perpendicular, or neither.\r\n\r\n[reveal-answer q=\"981152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981152\"]\r\n\r\nIdentify the slopes of the given lines.\r\n\r\nThe given lines are written in [latex]y=mx+b[\/latex] form, with [latex]m=\u22128[\/latex] for the first line and\u00a0[latex]m=\\frac{1}{8}[\/latex] for the second line.\r\n\r\nDetermine if the slopes are the same or if they are negative reciprocals.\r\n\r\n[latex]-8\\ne\\frac{1}{8}[\/latex], so the lines are not parallel.\r\n\r\nThe negative reciprocal of [latex]\u22128[\/latex] is [latex] \\displaystyle \\frac{1}{8}[\/latex], so the lines are perpendicular.\r\n\r\nThe slopes of the lines are negative reciprocals, so the lines are perpendicular.\r\n

      Answer<\/h4>\r\nThe lines are perpendicular.[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h2>\r\nhttps:\/\/youtu.be\/IIy4N2lAkDs\r\n\r\nIn this last example, you will work with an equation of a line given in Standard Form instead of Slope-Intercept Form as the previous examples have been given in.\r\n
      \r\n

      Example 18<\/h3>\r\nConsider the equation: [latex]8x-5y=10[\/latex]\r\n\r\nWhat is the slope of a line that is parallel to the above line?\r\n\r\nWhat is the slope of a line that is perpendicular to the above line?\r\n\r\n[reveal-answer q=\"390273\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"390273\"]\r\n\r\nFirst, we must write the equation in Slope-Intercept Form.\u00a0 To do this, we must solve for y.\r\n

      [latex]8x-5y=10[\/latex]<\/p>\r\nSubtract 8x from both sides of the equation\r\n

      [latex]\\begin{array}{r}8x-5y=10\\\\\\underline{\\,\\,\\,\\,-8x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-8x}\\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-5y=-8x+10\\end{array}[\/latex]<\/p>\r\nDivide each term by -5\r\n

      [latex]\\frac{-5y}{-5}=\\frac{-8x}{-5}+\\frac{10}{-5}[\/latex]<\/p>\r\n

      [latex]y=\\frac{8}{5}x-2[\/latex]<\/p>\r\nNow that the equation is in Slope-Intercept Form, we know the slope of this line is [latex]\\frac{8}{5}[\/latex].\r\n\r\nSince parallel lines have the same slope, we now know that the slope of a line parallel to\u00a0[latex]8x-5y=10[\/latex] is also\u00a0[latex]\\frac{8}{5}[\/latex].\r\n\r\nPerpendicular lines have slopes that are negative reciprocals of each other.\u00a0 Since the negative reciprocal of\u00a0[latex]\\frac{8}{5}[\/latex] is\u00a0[latex]-\\frac{5}{8}[\/latex], a line perpendicular to\u00a0[latex]8x-5y=10[\/latex] would have a slope of\u00a0[latex]-\\frac{5}{8}[\/latex].\r\n\r\nSolution:\u00a0<\/strong>\r\n\r\nA line parallel to\u00a0[latex]8x-5y=10[\/latex] would have a slope of\u00a0[latex]\\frac{8}{5}[\/latex].\r\n\r\nA line perpendicular to\u00a0[latex]8x-5y=10[\/latex] would have a slope of\u00a0[latex]-\\frac{5}{8}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      Graph a line using the slope and y-intercept<\/h2>\r\nPreviously we learned how to graph linear equations by finding ordered pairs that are solutions to the equation and then using those points to create a graph of the line.\u00a0 Another option for graphing linear functions is using the slope and y-intercept.\r\n\r\nLet us consider the following function.\r\n

      [latex]f\\left(x\\right)=\\dfrac{1}{2}x+1[\/latex]<\/p>\r\nThe function is in slope-intercept form.\u00a0 Therefore, we know the slope of the line is [latex]\\dfrac{1}{2}[\/latex] and the [latex]y[\/latex]-intercept is the point [latex](0,1)[\/latex].\u00a0 Next, we will look at the slope.\r\n

      [latex]slope=\\frac{rise}{run}=\\frac{1}{2}[\/latex]<\/p>\r\nSo, starting from the [latex]y[\/latex]-intercept [latex](0,1)[\/latex], we can \"rise\" 1 unit and \"run\" 2 units to locate a second point on our line.\u00a0 In other words, we will go up 1 unit and to the right 2 units.\u00a0 That brings us to the point [latex](2,2)[\/latex].\u00a0 We can repeat that process from the point [latex](2,2)[\/latex] to locate an additional point on the line.\u00a0 Drawing a line through those points creates the graph of our equation.\r\n\r\n\"graph\r\n

      \r\n

      How To: Given the equation for a linear function, graph the function using the y<\/em>-intercept and slope<\/h3>\r\n
        \r\n \t
      1. Evaluate the function at an input value of zero to find the y-<\/em>intercept.\u00a0 This will always be [latex](0,b)[\/latex].<\/li>\r\n \t
      2. Identify the slope.<\/li>\r\n \t
      3. Plot the point represented by the y-<\/em>intercept.<\/li>\r\n \t
      4. Use [latex]\\dfrac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\r\n \t
      5. Sketch the line that passes through the points.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Example 19<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\dfrac{2}{3}x+5[\/latex] using the y-<\/em>intercept and slope.\r\n\r\n[reveal-answer q=\"421669\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"421669\"]\r\n\r\nThe function is written in slope-intercept form.\u00a0 Therefore, we know the slope of the line is [latex]-\\frac{2}{3}[\/latex] and the [latex]y[\/latex]-intercept is the point [latex](0,5)[\/latex].\u00a0 We will begin graphing by plotting the point [latex](0,5)[\/latex].\u00a0 From this point we will use the slope to locate a second point on the line.\u00a0 The fraction [latex]-\\frac{2}{3}[\/latex] can also be written as [latex]\\frac{-2}{3}[\/latex].\u00a0 Let's consider what this means in terms of our slope.\r\n

        [latex]slope=\\frac{rise}{run}=\\frac{-2}{3}[\/latex]<\/p>\r\nWhat do you think the negative in the numerator means in terms of locating our second point?\u00a0 When the numerator is positive, we go up the indicated number of units.\u00a0 Since our numerator is negative, we will go down 2 units instead.\u00a0 Going down 2 units and then right 3 units brings us to the point [latex](3,3)[\/latex].\u00a0 Drawing a line through the points [latex](0,5)[\/latex] and [latex](3,3)[\/latex] creates the graph of this equation.\r\n\r\n\"Coordinate\r\n\r\nNotice that the line slants downward from left to right.\u00a0 This is consistent with having a negative slope.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show another example of how to graph a linear function given the y-intercepts and the slope.\r\n\r\nhttps:\/\/youtu.be\/N6lEPh11gk8\r\n\r\nIn the next example, we will graph another linear function using the slope and [latex]y[\/latex]-intercept.\u00a0 However, in this example, the equation is not given in slope-intercept form.\r\n

        \r\n

        Example 20<\/h3>\r\nGraph [latex]3x+4y=24[\/latex]\u00a0using the slope and y-intercept.\r\n[reveal-answer q=\"930515\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"930515\"]\r\n\r\nThis equation is not written in slope-intercept form.\u00a0 Therefore, the first thing we need to do is solve for [latex]y[\/latex] and rewrite the equation in slope-intercept form.\r\n

        [latex]3x+4y=24[\/latex]<\/p>\r\n

        [latex]\\underline{-3x\\hspace{.42in}-3x}\\hspace{.1in}[\/latex]<\/p>\r\n

        [latex]\\hspace{.95in}4y=-3x+24[\/latex]<\/p>\r\n

        [latex]\\frac{4y}{4}=\\frac{-3x}{4}+\\frac{24}{4}[\/latex]<\/p>\r\n

        [latex]y=-\\frac{3}{4}x+6[\/latex]<\/p>\r\nNow that the equation is in slope-intercept form, we know that the slope of the line is [latex]-\\frac{3}{4}[\/latex] and the [latex]y[\/latex]-intercept is the point [latex](0,6)[\/latex].\r\n\r\nTo graph the equation, we will first plot the point [latex](0,6)[\/latex].\u00a0 Then, using the slope to locate a second point on the line, we will go down 3 units (because the slope is negative) and then right 4 units.\u00a0 That brings us to the point [latex](4,3)[\/latex].\u00a0 Drawing a line through our two points gives us the graph of the equation.\r\n\r\n \r\n\r\n\"A\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        Summary<\/h2>\r\nSlope describes the steepness of a line. The slope of any line remains constant along the line. The slope can also tell you information about the direction of the line on the coordinate plane. Slope can be calculated either by looking at the graph of a line or by using the coordinates of any two points on a line. There are two common formulas for slope: [latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex] and [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex] where [latex]m=\\text{slope}[\/latex]\u00a0and [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex] and [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex] are two points on the line.\r\n\r\nThe images below summarize the slopes of different types of lines.\r\n\r\n\"Four\r\n

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

        section 3.4 Learning Objectives<\/h3>\n

        3.4: Slope of a Line<\/strong><\/p>\n

          \n
        • Identify the slope of a line from the graph<\/li>\n
        • Find the slope of a line given two points on a line<\/li>\n
        • Find the slope and y-intercept given the equation of a line<\/li>\n
        • Find the slope of horizontal and vertical lines<\/li>\n
        • Find the slopes of parallel and perpendicular lines<\/li>\n
        • Graph a line using the slope and y-intercept<\/li>\n<\/ul>\n<\/div>\n

           <\/p>\n

          Identify slope from a graph<\/h2>\n

          The mathematical definition of slope<\/b> is very similar to our everyday one. In math, slope is used to describe the steepness and direction of lines. By just looking at the graph of a line, you can learn some things about its slope, especially relative to other lines graphed on the same coordinate plane. Consider the graphs of the three lines shown below:<\/p>\n

          \"Coordinate<\/p>\n

          First, let\u2019s look at lines A and B. If you imagined these lines to be hills, you would say that line B is steeper than line A. Line B has a greater slope than line A.<\/p>\n

          Next, notice that lines A and B slant up as you move from left to right. We say these two lines have a positive slope. Line C slants down from left to right. Line C has a negative slope. Using two of the points on the line, you can find the slope of the line by finding the rise and the run. The vertical change between two points is called the rise<\/b>, and the horizontal change is called the run<\/b>. The slope equals the rise divided by the run: [latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/p>\n

          \"A<\/p>\n

          You can determine the slope of a line from its graph by looking at the rise and run. One characteristic of a line is that its slope is constant all the way along it. So, you can choose any 2 points along the graph of the line to figure out the slope. Let\u2019s look at an example.<\/p>\n

          \n

          Example 1<\/h3>\n

          Use the graph to find the slope of the line.<\/p>\n

          \"A<\/b><\/p>\n

          Show Solution<\/span><\/p>\n
          \n\n\n\n\n\n
          [latex]\\text{rise}=2[\/latex]<\/td>\nStart from a point on the line, such as [latex](2,1)[\/latex] and move vertically until in line with another point on the line, such as [latex](6,3)[\/latex]. The rise is 2 units. It is positive because you moved up.<\/td>\n<\/tr>\n
          [latex]\\text{run}=4[\/latex]<\/td>\nNext, move horizontally to the point [latex](6,3)[\/latex]. Count the number of units. The run is 4 units. It is positive because you moved to the right.<\/td>\n<\/tr>\n
          [latex]\\displaystyle \\text{Slope}=\\frac{2}{4}=\\frac{1}{2}[\/latex]<\/td>\n[latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Answer<\/h4>\n

          [latex]Slope=\\frac{1}{2}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n

          This line will have a slope of [latex]\\displaystyle \\frac{1}{2}[\/latex] no matter which two points you pick on the line. Try measuring the slope from the origin, [latex](0,0)[\/latex], to the point [latex](6,3)[\/latex]. You will find that the [latex]\\text{rise}=3[\/latex] and the [latex]\\text{run}=6[\/latex]. The slope is [latex]\\displaystyle \\frac{\\text{rise}}{\\text{run}}=\\frac{3}{6}=\\frac{1}{2}[\/latex]. It is the same!<\/p>\n

          Let\u2019s look at another example.<\/p>\n

          \n

          Example 2<\/h3>\n

          Use the graph to find the slope of the two lines.\u00a0<\/b><\/p>\n

          \"Two<\/b><\/p>\n

          Show Solution<\/span><\/p>\n
          Notice that both of these lines have positive slopes, so you expect your answers to be positive.<\/p>\n\n\n\n\n\n\n\n\n\n\n
          Blue line<\/th>\n<\/tr>\n
          [latex]\\text{rise}=4[\/latex]<\/td>\nStart with the blue line, going from point [latex](-2,1)[\/latex] to point [latex](-1,5)[\/latex]. This line has a rise of 4 units up, so it is positive.<\/td>\n<\/tr>\n
          [latex]\\text{run}=1[\/latex]<\/td>\nRun is 1 unit to the right, so it is positive.<\/td>\n<\/tr>\n
          [latex]\\displaystyle \\text{Slope }=\\frac{4}{1}=4[\/latex]<\/td>\nSubstitute the values for the rise and run in the formula [latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/td>\n<\/tr>\n
          Red line<\/th>\n<\/tr>\n
          [latex]\\text{rise}=1[\/latex]<\/td>\nThe red line, going from point [latex](-1,-2)[\/latex] to point [latex](3,-1)[\/latex] has a rise of 1 unit.<\/td>\n<\/tr>\n
          [latex]\\text{run}=4[\/latex]<\/td>\nThe red line has a run of 4 units.<\/td>\n<\/tr>\n
          [latex]\\displaystyle \\text{Slope }=\\frac{1}{4}[\/latex]<\/td>\nSubstitute the values for the rise and run into the formula [latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Answer<\/h4>\n

          The slope of the blue line is 4 and the slope of the red line is [latex]\\frac{1}{4}[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n

          When you look at the two lines, you can see that the blue line is steeper than the red line. It makes sense the value of the slope of the blue line, 4, is greater than the value of the slope of the red line, [latex]\\displaystyle \\frac{1}{4}[\/latex]. The greater the slope, the steeper the line.<\/p>\n