{"id":6698,"date":"2018-03-16T15:55:30","date_gmt":"2018-03-16T15:55:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/?post_type=chapter&p=6698"},"modified":"2018-06-04T15:27:17","modified_gmt":"2018-06-04T15:27:17","slug":"slope-and-equations-of-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-albany-chemistry\/chapter\/slope-and-equations-of-lines\/","title":{"raw":"0.9 Slope and Equations of Lines","rendered":"0.9 Slope and Equations of Lines"},"content":{"raw":"
\r\n\r\n[embed]https:\/\/youtu.be\/A_V_J9yXi3g[\/embed]\r\n

QUICK reference<\/h3>\r\n

[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.<\/p>\r\n \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<\/div>\r\n

Identify slope from a graph<\/h2>\r\nIn 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\"Three\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<\/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\"]Start 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 as you moved up.\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 as 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 as 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]\\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<\/h3>\r\nUse the graph to find the slope of the two lines.\u00a0<\/b>\r\n\r\n\"A<\/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

Distinguish between graphs of lines with negative and positive slopes<\/h2>\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 (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<\/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\nStart at Point A, [latex](0,4)[\/latex] and rise [latex]\u22123[\/latex]. This means moving 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{\\text{rise}}{\\text{run}}[\/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 found the slope by starting at point B, running [latex]{-2}[\/latex], and then rising [latex]+3[\/latex] 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

Finding the Slope from\u00a0Two 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 straight line without its graph if you know the coordinates of any two points on that line. Every point has a set of coordinates: an 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 distance 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\"A\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 [latex]\\left(y_{2},y_{1}\\right)[\/latex]\u00a0<\/sub>and [latex]\\left(x_{2},x_{1}\\right)[\/latex].\r\n
\r\n

Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](5,5)[\/latex] and [latex](4,2)[\/latex]?\r\n\r\n[reveal-answer q=\"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<\/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 (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\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

Finding the Slopes 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\"The\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 their case, no matter which two points you choose, they will always have the same x<\/i>-coordinate. The equation for this line is [latex]x=2[\/latex].\r\n\r\n\"The\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\\left(x=0y+2\\right)[\/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 has no meaning for the set of real numbers. Because of this fact, it is said that the slope of this vertical line is undefined. This is true for all vertical lines\u2014they all have a slope that is undefined.\r\n

\r\n

Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](3,2)[\/latex] and [latex](\u22128,2)[\/latex]?\r\n\r\n[reveal-answer q=\"111566\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111566\"]\r\n\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\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
    \r\n

    Example<\/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<\/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
    \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 4, then the slope of the second equation will need to be [latex]-\\frac{1}{4}[\/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] 4\\cdot-\\frac{1}{4}=\\frac{4}{1}\\cdot-\\frac{1}{4}=-1[\/latex].\r\n
    \r\n

    Example<\/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 is 2.\r\n\r\nIdentify the slope of the given line.\r\n

    Answer<\/h4>\r\nThe slope of the perpendicular line is [latex]-\\tfrac{1}{2}[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nTo find the slope of a perpendicular line, find the reciprocal, [latex] \\displaystyle \\tfrac{1}{2}[\/latex], and then find the opposite of this reciprocal [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 slopes 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<\/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 opposite reciprocals.\r\n\r\n[latex]-8\\ne\\frac{1}{8}[\/latex], so the lines are not parallel.\r\n\r\nThe opposite 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 opposite 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\n

    Verify Slope From a Dataset<\/h2>\r\nMassive amounts of data is being collected every day by a wide range of institutions and groups. \u00a0This data is used for many purposes including business decisions about location and marketing, government decisions about allocation of resources and infrastructure, and personal decisions about where to live or where to buy food.\r\n\r\nIn the following example, you will see how a dataset can be used to define\u00a0the slope of a linear equation.\r\n
    \r\n

    Example<\/h3>\r\nGiven the dataset, verify the values of the slopes of each equation.\r\n\r\nLinear equations describing the change in median home values between 1950 and 2000 in Mississippi and Hawaii are as follows:\r\n\r\nHawaii:\u00a0<\/strong> [latex]y=3966x+74,400[\/latex]\r\n\r\nMississippi:\u00a0\u00a0<\/strong>[latex]y=924x+25,200[\/latex]\r\n\r\nThe equations are based on the following dataset.\r\n\r\nx = the number of years since 1950, and y = the median value of a house in the given state.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Year (x<\/em>)<\/th>\r\nMississippi House Value (y<\/em>)<\/th>\r\nHawaii House Value (y<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    0<\/td>\r\n$25,200<\/td>\r\n$74,400<\/td>\r\n<\/tr>\r\n
    50<\/td>\r\n$71,400<\/td>\r\n$272,700\u00a0\u00a0<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe slopes of each equation can be calculated with the formula you learned in the section on slope.\r\n

    [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex]<\/p>\r\nMississippi:<\/strong>\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(0, 25,200)<\/td>\r\n[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=25,200\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n
    Point 2<\/td>\r\n(50, 71,400)<\/td>\r\n[latex]\\begin{array}{l}x_{2}=50\\\\y_{2}=71,400\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    [latex] \\displaystyle m=\\frac{{71,400}-{25,200}}{{50}-{0}}=\\frac{{46,200}}{{50}} = 924[\/latex]<\/p>\r\nWe have verified that the slope [latex] \\displaystyle m = 924[\/latex] matches the dataset provided.\r\n\r\nHawaii:<\/strong>\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(0, 74,400)<\/td>\r\n[latex]\\begin{array}{l}x_{1}=1950\\\\y_{1}=74,400\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n
    Point 2<\/td>\r\n(50, 272,700)<\/td>\r\n[latex]\\begin{array}{l}x_{2}=2000\\\\y_{2}=272,700\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    [latex]\\displaystyle m=\\frac{{272,700}-{74,400}}{{50}-{0}}=\\frac{{198,300}}{{50}} = 3966[\/latex]<\/p>\r\nWe have verified that the slope [latex] \\displaystyle m = 3966[\/latex] matches the dataset provided.\r\n\r\n<\/div>\r\n

    \r\n

    Example<\/h3>\r\nGiven the dataset, verify the values of the slopes of the\u00a0equation.\r\n\r\nA linear equation describing the change in the number of high school students who smoke, in\u00a0a group of 100, between 2011 and 2015 is given as:\r\n

    \u00a0[latex]y = -1.75x+16[\/latex]<\/p>\r\nAnd is based on the data from this table, provided by the Centers for Disease Control.\r\n\r\nx = the number of years since 2011, and y = the number of high school smokers per 100 students.\r\n\r\n\r\n\r\n\r\n\r\n
    Year<\/td>\r\nNumber of \u00a0High School Students Smoking\u00a0Cigarettes (per 100)<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\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(0, 16)<\/td>\r\n[latex]\\begin{array}{l}x_{1}=0\\\\y_{1}=16\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n
    Point 2<\/td>\r\n(4, 9)<\/td>\r\n[latex]\\begin{array}{l}x_{2}=4\\\\y_{2}=9\\end{array}[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

    [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\\frac{{9-16}}{{4-0}} =\\frac{{-7}}{{4}}=-1.75[\/latex]<\/p>\r\nWe have verified that the slope [latex] \\displaystyle{m=-1.75}[\/latex] matches the dataset provided.\r\n\r\n<\/div>\r\n

    Interpret the Slope of \u00a0Linear Equation<\/h2>\r\nOkay, now we have verified that data can provide us with the slope of a linear equation. So what? We can use this information to describe how something changes using words.\r\n\r\nFirst, let's review the different kinds of slopes possible in a linear equation.\r\n\r\n\"Uphill\r\n\r\nWe often use specific words to describe the different types of slopes when we are using lines and equations to represent \"real\" situations. The following table pairs the type of slope with the common language used to describe it both verbally and visually.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Type of Slope<\/strong><\/td>\r\nVisual Description\u00a0<\/strong><\/td>\r\nVerbal Description<\/strong><\/td>\r\n<\/tr>\r\n
    positive<\/td>\r\nuphill<\/td>\r\nincreasing<\/td>\r\n<\/tr>\r\n
    negative<\/td>\r\ndownhill<\/td>\r\ndecreasing<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\nhorizontal<\/td>\r\nconstant<\/td>\r\n<\/tr>\r\n
    undefined<\/td>\r\nvertical<\/td>\r\nN\/A<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
    \r\n

    Example<\/h3>\r\nInterpret the slope of each equation for house values using words.\r\n\r\nHawaii:\u00a0<\/strong> [latex]y = 3966x+74,400[\/latex]\r\n\r\nMississippi:\u00a0\u00a0<\/strong>[latex]y = 924x+25,200[\/latex]\r\n\r\n[reveal-answer q=\"871726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"871726\"]It helps to apply the units to the points that we used to define slope. \u00a0The x<\/em>-values represent years, and the y<\/em>-values represent dollar amounts.\r\n\r\nFor Mississippi:\r\n

    [latex] \\displaystyle m=\\frac{{71,400}-{25,200}}{{0}-{50}}=\\frac{{46,200\\text{ dollars}}}{{50\\text{ year}}} = 924\\frac{\\text{dollars}}{\\text{year}}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe slope for the Mississippi home prices equation is positive<\/strong>, so each year the price of a home in Mississippi\u00a0increases<\/strong> by 924 dollars.\r\n\r\nWe can apply the same thinking for Hawaii home prices. The slope for the Hawaii\u00a0home prices equation tells us that each year, the price of a home increases by 3966\u00a0dollars.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example<\/h3>\r\nInterpret the slope of the line describing the change in the number of high school smokers using words.\r\n\r\nApply units to the formula for slope. The x<\/em> values represent years, and the y<\/em> values represent the number of smokers. Remember that this dataset is per 100 high school students.\r\n

    [latex] \\displaystyle m=\\frac{{9-16}}{{2015-2011}} =\\frac{{-7 \\text{ smokers}}}{{4\\text{ year}}}=-1.75\\frac{\\text{ smokers}}{\\text{ year}}[\/latex]<\/p>\r\nThe slope of this linear equation is negative<\/strong>, so this tells us that there is a decrease<\/strong> in the number of high school age smokers each year.\r\n\r\nThe number of high schoolers that smoke decreases by 1.75 per 100 each year.\r\n\r\n<\/div>\r\n \r\n

    Equations of Lines<\/h2>\r\nWhen graphing a line one method we could use is to make a table of values. However, if we can identify some properties of the line, we may be able to make a graph much quicker and easier. One such method is finding the slope and the y<\/em>-intercept of the equation. The slope can be represented by m and the y<\/em>-intercept, where it crosses the axis and [latex]x=0[\/latex], can be represented by [latex](0,b)[\/latex] where b<\/em> is the value where the graph crosses the vertical y<\/em>-axis. Any other point on the line can be represented by [latex](x,y)[\/latex].\r\n\r\nIn the equation,\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

    This formula is known as the slope-intercept equation.\u00a0If we know the slope and the y<\/em>-intercept we can easily find the equation that represents the line<\/p>\r\n\r\n

    \r\n

    Example<\/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]\u22125[\/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<\/h3>\r\nWrite the equation of the line in the graph by identifying the slope and y<\/em>-intercept.\r\n\"SVG_Grapher\"\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 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\nWe can also move the opposite direction, using the equation identify the slope\u00a0and y<\/em>-intercept and graph the equation from this information. However, it will be\u00a0important for the equation to first be in slope intercept form. If it is not, we will\u00a0have to solve it for y<\/em> so we can identify the slope and the y<\/em>-intercept.\r\n
    \r\n

    Example<\/h3>\r\nWrite the following equation in slope-intercept form.\r\n

    [latex]2x+4y=6[\/latex]<\/p>\r\n[reveal-answer q=\"373034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373034\"]We need to solve for y<\/em>. Start by subtracting [latex]2[\/latex] from both sides.\r\n

    [latex]\\begin{array}{r}2x\\,\\,\\,+\\,\\,\\,4y\\,\\,\\,=\\,\\,\\,6\\\\-2x\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,-2x\\end{array}[\/latex]<\/p>\r\n

    It helps to place the x<\/em> term first on the right hand side. Notice how we keep the 6 positive by placing an addition sign in front.<\/p>\r\n

    [latex]4y=-2x+6[\/latex]<\/p>\r\n

    Divide each term by 4 to isolate the y<\/em>.<\/p>\r\n

    [latex]\\frac{4y}{4}=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\r\n

    [latex]y=\\frac{-2x}{4}+\\frac{6}{4}[\/latex]<\/p>\r\n

    Reduce the fractions<\/p>\r\n

    [latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]y=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOnce we have an equation in slope-intercept form we can graph it by first plotting\u00a0the y<\/em>-intercept, then using the slope, find a second point and connecting the dots.\r\n
    \r\n

    Example<\/h3>\r\nGraph [latex]y=\\frac{1}{2}x-4[\/latex] using the slope-intercept equation.\r\n\r\n[reveal-answer q=\"420487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"420487\"]First, plot the y<\/em>-intercept.\r\n\r\n\"The\r\n\r\nNow use the slope to count up or down and over left or right to the next point. This slope is [latex]\\frac{1}{2}[\/latex], so you can count up one and right two\u2014both positive because both parts of the slope are positive.\r\n\r\nConnect the dots.\r\n\"A\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n

    Find the Equation of a Line Given the Slope and a Point on the Line<\/h2>\r\nUsing the slope-intercept equation of a line is possible 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\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\nThis may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.\r\n
    \r\n

    Example<\/h3>\r\nWrite the equation of the line that has a slope of 3 and contains the point [latex](1,4)[\/latex].\r\n\r\n[reveal-answer q=\"161353\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161353\"]\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][\/hidden-answer]\r\n\r\n<\/div>\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\nIf you know the slope of a line and a point on the line, you can draw a graph. Using an equation in the point-slope form allows you to identify the slope and a point. Consider the equation [latex] \\displaystyle y=-3x-1[\/latex]. The y<\/em>-intercept is the point on the line where it passes through the y<\/em>-axis. What is the value of x<\/em> at this point?\r\n
    Reminder: All y<\/em>-intercepts are points in the form [latex](0,y)[\/latex]. \u00a0The x<\/em> value of any y<\/em>-intercept is always<\/em>\u00a0zero.<\/div>\r\nTherefore, you can tell from this equation that the y<\/i>-intercept is at [latex](0,\u22121)[\/latex], check this by replacing x<\/em> with 0 and solving for y<\/em>. To graph the line, start by plotting that point, [latex](0,\u22121)[\/latex], on a graph.\r\n\r\nYou can also tell from the equation that the slope of this line is [latex]\u22123[\/latex]. So start at [latex](0,\u22121)[\/latex] and count up 3 and over [latex]\u22121[\/latex] (1 unit in the negative direction, left) and plot a second point. (You could also have gone down 3 and over 1.) Then draw a line through both points, and there it is, the graph of [latex] \\displaystyle y=-3x-1[\/latex].\r\n\r\n\"A\r\n
    \r\n

    Example (Advanced)<\/h3>\r\nWrite the equation of the line that has a slope of [latex]-\\frac{7}{8}[\/latex]\u00a0and contains the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"31452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"31452\"]\r\n\r\nSubstitute the slope (m<\/i>) into [latex]y=mx+b[\/latex].\r\n

    [latex]\\begin{array}{l}y=mx+b\\\\\\\\y=-\\frac{7}{8}x+b\\end{array}[\/latex]<\/p>\r\nSubstitute the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex]\u00a0for x <\/i>and y.<\/i>\r\n

    [latex]\\frac{5}{4}=-\\frac{7}{8}\\left(4\\right)+b[\/latex]<\/p>\r\nSolve for b.<\/i>\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{28}{8}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{14}{4}+b\\\\\\\\\\frac{5}{4}+\\frac{14}{4}=-\\frac{14}{4}+\\frac{14}{4}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{19}{4}=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex] with [latex] \\displaystyle m=-\\frac{7}{8}[\/latex] and [latex] \\displaystyle b=\\frac{19}{4}[\/latex].\r\n

    Answer<\/h4>\r\n[latex]y=-\\frac{7}{8}x+\\frac{19}{4}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n

    Find 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. You will again use slope-intercept form to help you.\r\n\r\nThe slope of a linear equation is always the same, no matter which two points you use to find the slope. Since you have two points, you can use those points to find the slope (m<\/i>). Now you have the slope and a point on the line! You can now substitute values for m<\/i>, x<\/i>, and y<\/i> into the equation [latex]y=mx+b[\/latex] and find b<\/em>.\r\n
    \r\n

    Example<\/h3>\r\nWrite the equation of the line that passes through the points [latex](2,1)[\/latex] and [latex](\u22121,\u22125)[\/latex].\r\n\r\n[reveal-answer q=\"333536\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"333536\"]\r\n\r\nFind 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 (m<\/i>) into [latex]y=mx+b[\/latex].\r\n

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

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

    Answer<\/h4>\r\n[latex]\\begin{array}{l}y=2x+\\left(-3\\right)\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\text{or}\\\\y=2x-3\\end{array}[\/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

    \r\n

    Example (Advanced)<\/h3>\r\nWrite the equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex].\r\n\r\n[reveal-answer q=\"347882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"347882\"]\r\n\r\nFind the slope using the given points.\r\n

    [latex] \\displaystyle \\frac{7.6-6.45}{1.15-(-4.6)}=\\frac{1.15}{5.75}=0.2[\/latex]<\/p>\r\nSubstitute the slope (m<\/i>) into [latex] \\displaystyle y=mx+b[\/latex].\r\n

    [latex] \\displaystyle y=0.2x+b[\/latex]<\/p>\r\nSubstitute either point for x <\/i>and y\u2014<\/i>this example uses [latex](1.15,7.6)[\/latex]. Then solve for b<\/i>.\r\n

    [latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.2(1.15)+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\,\\,\\,\\,\\,\\,7.6\\,\\,=\\,\\,0.23+b\\\\\\underline{-0.23\\,\\,\\,\\,-0.23}\\\\\\,\\,\\,\\,\\,7.37\\,=\\,\\,b\\end{array}[\/latex]<\/p>\r\nRewrite [latex] \\displaystyle y=mx+b[\/latex] with [latex]m=0.2[\/latex] and [latex]b=7.37[\/latex].\r\n

    [latex] \\displaystyle y=0.2x+7.37[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\nThe equation of the line that passes through the points [latex](-4.6,6.45)[\/latex] and [latex](1.15,7.6)[\/latex] is [latex]y=0.2x+7.37[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n

    Write the equations of parallel and perpendicular lines<\/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
    \r\n

    Example<\/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] and [latex]b=\u22125[\/latex].\r\n\r\nSince [latex]m=1[\/latex], 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. Substitute 1 for m<\/i>, and the point [latex](\u22122,1)[\/latex] for x<\/i> and y<\/em>.\r\n

    [latex]\\begin{array}{l}y=mx+b\\\\1=1(\u22122)+b\\end{array}[\/latex]<\/p>\r\nSolve for b<\/em>.\r\n

    [latex]\\begin{array}{l}1=\u22122+b\\\\3=b\\end{array}[\/latex]<\/p>\r\nWrite the equation using the new slope for m<\/i> and the b<\/i> you just found.\r\n

    Answer<\/h4>\r\n[latex]y=x+3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n

    Determine the Equation of a Line Perpendicular to Another Line Through a Given Point<\/h2>\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<\/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\nUse the method for writing an equation from the slope and a point on the line. Substitute [latex] \\displaystyle -\\frac{1}{2}[\/latex] for m<\/i>, and the point [latex](1,5)[\/latex] for x<\/i> and y<\/i>.\r\n

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

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

    Example<\/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<\/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

    Interpret the y<\/em>-intercept of a linear equation<\/h2>\r\nOften, when the line in question represents a set of data or observations, the y<\/em>-intercept can be interpreted as a starting point. \u00a0We will continue to use the examples for house value in Mississippi and Hawaii and high school smokers to interpret the meaning of the y<\/em>-intercept in those equations.\r\n
    \r\n

    Example<\/h3>\r\nRecall the equations and data for house value:<\/strong>\r\n\r\nLinear equations describing the change in median home values between 1950 and 2000 in Mississippi and Hawaii are as follows:\r\n\r\nHawaii:\u00a0<\/strong> [latex]y = 3966x+74,400[\/latex]\r\n\r\nMississippi:\u00a0\u00a0<\/strong>[latex]y = 924x+25,200[\/latex]\r\n\r\nThe equations are based on the following dataset.\r\n\r\nx = the number of years since 1950, and y = the median value of a house in the given state.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    Year (x<\/em>)<\/th>\r\nMississippi House Value (y<\/em>)<\/th>\r\nHawaii House Value (y<\/em>)<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
    0<\/td>\r\n$25,200<\/td>\r\n$74,400<\/td>\r\n<\/tr>\r\n
    50<\/td>\r\n$71,400<\/td>\r\n$272,700\u00a0\u00a0<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAnd the equations and data for high school smokers:\r\n\r\nA linear equation describing the change in the number of high school students who smoke, in\u00a0a group of 100, between 2011 and 2015 is given as:\r\n

    \u00a0[latex]y = -1.75x+16[\/latex]<\/p>\r\nAnd is based on the data from this table, provided by the Centers for Disease Control.\r\n\r\nx = the number of years since 2011, and y = the number of high school smokers per 100 students.\r\n\r\n\r\n\r\n\r\n\r\n
    Year<\/td>\r\nNumber of \u00a0High School Students Smoking\u00a0Cigarettes (per 100)<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAlso recall that the equation of a line in slope-intercept form is as follows:\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<\/div>\r\n

    The examples that follow show how to interpret the y-intercept of the equations used to model house value and the number of high school smokers. Additionally, you will see how to use the equations to make predictions about house value and the number of smokers in future years.<\/p>\r\n\r\n

    \r\n

    Example<\/h3>\r\nInterpret the y<\/em>-intercepts of the equations that represent the change in house value for Hawaii and Mississippi.\r\n\r\nHawaii:\u00a0<\/strong> [latex]y = 3966x+74,400[\/latex]\r\n\r\nMississippi:\u00a0\u00a0<\/strong>[latex]y = 924x+25,200[\/latex]\r\n\r\nThe y<\/em>-intercept of a two-variable linear equation can be found by substituting 0 in for x.\r\n

    Hawaii<\/h4>\r\n

    [latex]y = 3966x+74,400\\\\y = 3966(0)+74,400\\\\y = 74,400[\/latex]<\/p>\r\nThe y<\/em>-intercept is a point, so we write it as (0, 74,400). \u00a0Remember that y<\/em>-values represent dollars and x<\/em> values represent years. \u00a0When the year is 0\u2014in this case 0\u00a0because that is the first date we have in the dataset\u2014the price of a house in Hawaii was $74,400.\r\n

    Mississippi<\/h4>\r\n

    [latex]y = 924x+25,200\\\\y = 924(0)+25,200\\\\y = 25,200[\/latex]<\/p>\r\nThe y<\/em>-intercept is (0,\u00a025,200). \u00a0This means that in 1950 the value of a house in Mississippi was $25,200. Remember that x<\/em> represents the number of years since 1950, so if [latex]x=0[\/latex] the year is 1950.\r\n\r\n<\/div>\r\n

    <\/h2>\r\n
    \r\n

    Example<\/h3>\r\nInterpret the y-intercept of the equation that represents the change in the number of high school students who smoke out of 100.\r\n\r\nSubstitute 0 in for x<\/em>.\r\n

    [latex]y = -1.75x+16\\\\y = -1.75(0)+16\\\\y = 16[\/latex]<\/p>\r\nThe y-intercept is [latex](0,16)[\/latex]. \u00a0The data starts at 2011, so we represent that year as 0. We can interpret the y<\/em>-intercept as follows:\r\n\r\nIn the year 2011, 16 out of every 100 high school students smoked.\r\n\r\n<\/div>\r\n

    Use a linear equation to make a prediction<\/h2>\r\nAnother useful outcome we gain from writing equations from data is the ability to make predictions about what may happen in the future. We will continue our analysis of the house price and high school smokers. In the following examples you will be shown how to predict future outcomes based on the linear equations that model current behavior.\r\n
    \r\n

    Example<\/h3>\r\nUse the equations for house value in Hawaii and Mississippi to predict house value in\u00a02035.\r\n\r\nWe are asked to find house value, y<\/em>, when the year, x<\/em>, is 2035. Since the equations we have represent house value increase since 1950, we have to be careful. We can't just plug in 2035 for x<\/em>, because x<\/em> represents the years since 1950.\r\n\r\nHow many years are between 1950 and 2035? [latex]2035 - 1950 = 85[\/latex]\r\n\r\nThis is our x<\/em>-value.\r\n\r\nFor Hawaii:\r\n

    [latex]y = 3966x+74,400\\\\y = 3966(85)+74,400\\\\y = 337110+74,400 = 411,510[\/latex]<\/p>\r\nHoly cow! The average price for a house in Hawaii in 2035 is predicted to be $411,510 according to this model. See if you can find the current<\/em> average value of a house in Hawaii. Does the model measure up?\r\n\r\nFor Mississippi:\r\n

    [latex]y = 924x+25,200\\\\y = 924(85)+25,200\\\\y = 78540+25,200 = 103,740[\/latex]<\/p>\r\nThe average price for a home in Mississippi in 2035 is predicted to be $103,740 according to the model.\u00a0See if you can find the current<\/em> average value of a house in Mississippi. Does the model measure up?\r\n\r\n<\/div>\r\n

    \r\n

    Example<\/h3>\r\nUse the equation for the number of high school smokers per 100 to predict the year when there will be 0 smokers per 100.\r\n

    [latex]y = -1.75x+16[\/latex]<\/p>\r\nThis question takes a little more thinking. \u00a0In terms of x<\/em> and y<\/em>, what does it mean to have 0 smokers? \u00a0Since y<\/em> represents the number of smokers and x<\/em> represent the year, we are being asked when y will be 0.\r\n\r\nSubstitute 0 for y<\/em>.\r\n

    [latex]y = -1.75x+16[\/latex]<\/p>\r\n

    [latex]0 = -1.75x+16[\/latex]<\/p>\r\n

    [latex]-16 = -1.75x[\/latex]<\/p>\r\n

    [latex]\\frac{-16}{-1.75} = x[\/latex]<\/p>\r\n

    [latex]x = 9.14[\/latex] years<\/p>\r\nAgain, like the last example,\u00a0x<\/em> is representing the number of years since the start of the data\u2014which was 2011, based on the table:\r\n\r\n\r\n\r\n\r\n\r\n
    Year<\/td>\r\nNumber of \u00a0High School Students Smoking\u00a0Cigarettes (per 100)<\/td>\r\n<\/tr>\r\n
    0<\/td>\r\n16<\/td>\r\n<\/tr>\r\n
    4<\/td>\r\n9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo we are predicting that there will be no smokers in high school by [latex]2011+9.14=2020[\/latex]. How accurate do you think this model is? Do you think there will ever be 0 smokers in high school?\r\n\r\n<\/div>\r\n

    Bringing it Together<\/h2>\r\nThe last example we will show will include many\u00a0of the concepts that we have been building up throughout this section. \u00a0We will interpret a word problem, write a linear equation from it, graph the equation, interpret the y-intercept and make a prediction. Hopefully this example will help you to make\u00a0connections between the concepts we have presented.\r\n
    \r\n

    Example<\/h3>\r\nIt costs $600 to purchase an iphone, plus $55 per month for unlimited use and data.\r\n\r\nWrite a linear equation that represents the cost, y, \u00a0of owning and using the\u00a0iPhone for x amount of months. When you have written your equation, answer the following questions:\r\n
      \r\n \t
    1. What is the total cost you\u2019ve paid after\u00a0owning and using your phone for 24 months?<\/li>\r\n \t
    2. If you have spent\u00a0$2,580 since you purchased your phone, how many months have you used your phone?<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_4649\" align=\"alignnone\" width=\"206\"]\"5 iPhone[\/caption]\r\n\r\n[reveal-answer q=\"282349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"282349\"][\/hidden-answer]\r\n\r\nRead and Understand:<\/strong>\u00a0We need to write a linear equation that represents the cost of owning and using an iPhone for any number of months. \u00a0We are to use y to represent cost, and x to represent the number of months we have used the phone.\r\n\r\nDefine and Translate:\u00a0<\/strong>We will use the slope-intercept form of a line, [latex]y=mx+b[\/latex], because we are given a starting cost and a monthly cost for use. \u00a0We will need to find the slope and the y-intercept.\r\n\r\nSlope: in this case we don't know two points, but we are given a rate in dollars for monthly use of the phone. \u00a0Our units are dollars per month because slope is [latex]\\frac{\\Delta{y}}{\\Delta{x}}[\/latex], and y is in dollars and x is in months. The slope will be [latex]\\frac{55\\text{ dollars }}{1\\text{ month }}[\/latex]. [latex]m=\\frac{55}{1}=55[\/latex]\r\n\r\nY-Intercept: the y-intercept is defined as a point [latex]\\left(0,b\\right)[\/latex]. \u00a0We want to know how much money we have spent, y, after 0 months. \u00a0We haven't paid for service yet, but we have paid $600 for the phone. The y-intercept in this case is called an initial cost. [latex]b=600[\/latex]\r\n\r\nWrite and Solve:\u00a0<\/strong>Substitute the slope and intercept you defined into the slope=intercept equation.\r\n

      [latex]\\begin{array}{c}y=mx+b\\\\y=55x+600\\end{array}[\/latex]<\/p>\r\n

      Now we will answer the following questions:<\/p>\r\n\r\n

        \r\n \t
      1. What is the total cost you\u2019ve paid after\u00a0owning and using your phone for 24 months?<\/li>\r\n<\/ol>\r\nSince x represents the number of months you have used the phone, we can substitute x=24 into our equation.\r\n

        [latex]\\begin{array}{c}y=55x+600\\\\y=55\\left(24\\right)+600\\\\y=1320+600\\\\y=1920\\end{array}[\/latex]<\/p>\r\n

        Y represents the cost after x number of months, so in this scenario, after 24 months, you have spent $1920 to own and use an iPhone.<\/p>\r\n\r\n

          \r\n \t
        1. If you have spent\u00a0$2,580 since you purchased your phone, how many months have you used your phone?<\/li>\r\n<\/ol>\r\nWe know that y represents cost, and we are given a cost and asked to find the number of months related to having spent that much. We will substitute y=$2,580 into the equation, then use what we know about solving linear equations to isolate x:\r\n

          \u00a0[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=55x+600\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,2580=55x+600\\\\\\text{ subtract 600 from each side}\\,\\,\\,\\,\\,\\,\\,\\underline{-600}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\underline{-600}\\\\\\text{}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,1980=55x\\\\\\text{}\\\\\\text{ divide each side by 55 }\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{1980}{55}=\\frac{55x}{55}\\\\\\text{}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,36=x\\end{array}[\/latex]<\/p>\r\n

          If you have spent $2,580 then you have been using your iPhone for 36 months, or 3 years.\u00a0<\/span><\/p>\r\n\r\n<\/div>","rendered":"

          \n