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2.2: Equations of lines

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    45034
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    In this section, we discuss applying formulas to obtain equations of lines, graphing lines, and special cases. If we can identify some properties of the line, we may be able to graph the line much quicker and easier.

    The Slope-Intercept Formula

    One such method is finding the slope and the \(y\)-intercept of the equation. The slope can be represented by \(m\), and the \(y\)-intercept can be represented by \((0, b)\), where \(b\) is the \(y\)-value at which the graph crosses the \(y\)-axis. Any other point on the line can be represented by \((x, y)\).

    Example \(\PageIndex{1}\)

    Prove the slope-intercept formula by using the slope formula for the line that has slope \(m\), \(y\)-intercept \((0, b)\), and passes through the point \((x, y)\).

    Solution

    \[\begin{array}{rl}m=\frac{y-b}{x-0}&\text{Simplify} \\ m=\frac{y-b}{x}&\text{Multiply both sides by }x \\ mx=y-b&\text{Add }b\text{ to both sides} \\ mx+b=y \\ y=mx+b&\text{Slope-intercept formula}\end{array}\nonumber\]

    Slope-Intercept Formula

    The slope-intercept formula of a linear equation is given by \(y = mx + b\), where \(m\) is the slope and \((0, b)\) is the \(y\)-intercept.

    Example \(\PageIndex{2}\)

    Find the equation of the line with slope \(\frac{3}{4}\) and \(y\)-intercept \((0, −3)\).

    Solution

    \[\begin{array}{rl}y=mx+b &m\text{ is the slope, }b\text{ is the }y\text{-intercept} \\ y=\color{blue}{\frac{3}{4}}\color{black}{}x\color{blue}{-3}\color{black}{}&\text{Equation of the line}\end{array}\nonumber\]

    Example \(\PageIndex{3}\)

    Find the equation of the line.

    clipboard_ea513535d29f11a6412f241edfbf252a2.png
    Figure \(\PageIndex{1}\)

    Solution

    The \(y\)-intercept is where the graph crosses the \(y\)-axis. We can see, from the graph, that the line crosses the \(y\)-axis at \((0, 3)\). Hence, \(b = 3\). To find the slope, we count the rise and run units. We can see that we rise downward \(2\) units and run to the right \(3\) units. Hence, the slope is \(−\frac{2}{3}\). Now we can put the equation of the line together in slope-intercept form, where \(m = −\frac{2}{3}\) and \(b = 3\):

    clipboard_eb11452329a886d6d3c64258600261da8.png
    Figure \(\PageIndex{2}\)

    \[y=-\frac{2}{3}x+3\nonumber\]

    Lines in Slope-Intercept Form

    We can also identify the slope and \(y\)-intercept, and graph the equation from a given equation. However, we need to be sure the equation is in slope-intercept form. If it is not, we will have to rewrite the equation in slope-intercept form, i.e., solve the equation for \(y\). Then we can easily identify the slope and the \(y\)-intercept.

    Example \(\PageIndex{4}\)

    Write the equation \(2x − 4y = 6\) in slope-intercept form. Find the slope and \(y\)-intercept of the line.

    Solution

    \[\begin{array}{rl}2x-4y=6&\text{Isolate the variable term }-4y \\ 2x-4y+\color{blue}{(-2x)}\color{black}{}=6+\color{blue}{(-2x)}\color{black}{}&\text{Simplify} \\ -4y=6-2x &\text{Multiply by the reciprocal of }-4 \\ \color{blue}{-\frac{1}{4}}\color{black}{}\cdot -4y=\color{blue}{-\frac{1}{4}}\color{black}{}\cdot 6-2x\cdot\color{blue}{-\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=\frac{1}{2}x-\frac{3}{2}&\text{Slope-intercept form}\end{array}\nonumber\]

    Next, we identify the slope and the \(y\)-intercept. If we line up the general slope-intercept form with the equation we obtained, we can easily see \(m\) and \(b\): \[\begin{array}{llll} y&=&mx&+b \\ y&=&\color{blue}{\frac{1}{2}}\color{black}{}x &\color{blue}{-\frac{3}{2}}\end{array}\nonumber\]

    Hence, \(m = \frac{1}{2}\) and \(b = −\frac{3}{2}\), i.e., the slope is \(\frac{1}{2}\) and the \(y\)-intercept is \(−\frac{3}{2}\).

    Graphing Lines

    Once we have an equation in slope-intercept form, we can graph it by first plotting the \(y\)-intercept, then applying the slope to find a second point and even a third point. We connect these points to make a line. Let’s look at Example \(\PageIndex{4}\) and graph the line.

    Example \(\PageIndex{5}\)

    Graph \(y =\frac{1}{2} x −\frac{3}{2}\) by using the slope and \(y\)-intercept.

    Solution

    The \(y\)-intercept, or \(b\), is where the graph crosses the \(y\)-axis. We know from Example \(\PageIndex{4}\) that the \(y\)-intercept is \(−\frac{3}{2}\) and the line will cross the \(y\)-axis at \(\left(0,-\frac{3}{2}\right)\). The slope is \(\frac{1}{2}\), and, using \(\frac{rise}{run}\), we need to rise upward \(1\) unit and run to the right \(2\) units to reach the next point. We continue the pattern to obtain a third point. Now we can connect the dots and create a well-defined line. Be sure to draw it to fill the grid.

    clipboard_e3081aa2630ae19555395a76a0a64b905.png
    Figure \(\PageIndex{3}\)
    Note

    When we graphed a line by plotting points from a table, we obtained three points. Hence, in Example \(\PageIndex{5}\), we obtained three points and then drew a line. Obtaining three points on a line is common practice and will help when drawing any line, even the special cases.

    Note

    Before our current system of graphing, French mathematician, Nicole Oresme, in 1323, suggested graphing lines that would look more like bar graphs with a constant slope.

    Example \(\PageIndex{6}\)

    Write the equation \(3x + 4y = 12\) in slope-intercept form. Find the slope and \(y\)-intercept of the line and then graph the line.

    Solution

    \[\begin{array}{rl}3x+4y=12&\text{Isolate the variable term }4y \\ 3x+4y+\color{blue}{(-3x)}\color{black}{}=12+\color{blue}{(-3x)}\color{black}{}&\text{Simplify} \\ 4y=12-3x&\text{Multiply by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4y=\color{blue}{\frac{1}{4}}\color{black}{}\cdot 12-3x\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=-\frac{3}{4}x+3&\text{Slope-intercept form}\end{array}\nonumber\]

    No we can graph the line. We see the \(y\)-intercept is \(3\) and the line will cross the \(y\)-axis at \((0, 3)\). The slope is \(−\frac{3}{4}\), and, using \(\frac{rise}{run}\), we need to rise downward \(3\) units and run to the right \(4\) units to reach the next point. We continue the pattern to obtain a third point. Now we can connect the dots and create a well-defined line. Be sure to draw it to fill the grid.

    clipboard_e4f1cb9e4b2d9b9957990a715bca9a9dc.png
    Figure \(\PageIndex{4}\)

    Vertical and Horizontal Lines

    Lines with zero or undefined slope can make a problem seem very different. Zero slope, or a horizontal line, will simply have a slope of zero. So, the equation simply becomes \(y = b\) or \(y\) equal to the \(y\)-coordinate of the graph. If we have undefined slope, or a vertical line, the equation can’t be written in slope-intercept form because the slope is undefined. Hence, there is no \(y\) in these equations. We will simply make \(x\) equal to the \(x\)-coordinate of the graph.

    Example \(\PageIndex{7}\)

    Graph the line \(x=-4\).

    Solution

    Since \(x = −4\) is a vertical line, then we know this line has no slope and the line is at its steepest. Every \(x\)-coordinate on this line is \(−4\) and the line has no run. We can graph this line easily by plotting three points where the \(x\)-coordinate is \(−4\). Let’s plot \((−4, −1),\: (−4, 0),\) and \((−4, 2)\); then connect the points with a well-defined line.

    clipboard_e2ad425d12646fa3b958050f71ecff582.png
    Figure \(\PageIndex{5}\)
    Example \(\PageIndex{8}\)

    Graph the line \(y=1\).

    Solution

    Since \(y = 1\) is a horizontal line, then we know this line has zero slope and the line is at its flattest. Every \(y\)-coordinate on this line is \(1\) and the line has no rise. We can graph this line easily by plotting three points where the \(y\)-coordinate is \(1\). Let’s plot \((3, 1),\: (0, 1),\) and \((−2, 1)\); then connect the points with a well-defined line.

    clipboard_e801ca06a8c2e1bab6a1daa9498b7a996.png
    Figure \(\PageIndex{6}\)

    Point-Slope Formula

    The slope-intercept form has the advantage of being simple to remember and use. However, it has one major disadvantage: we must know the \(y\)-intercept in order to graph the line. Generally, we do not know the \(y\)-intercept, but, usually, know one or more points on the line that are not the \(y\)-intercept. In these cases, we can’t use the slope-intercept equation, so we will need a more general formula to assist us in graphing lines. If the slope of a line is \(m\), and point \((x_1, y_1)\) be a particular point on the line, and any other point on the line be \((x, y)\), then we can use this to find this general formula.

    \[\begin{array}{rl}m,\: (x_1,y_1),\: (x,y)&\text{Recall slope formula} \\ \frac{y_2-y_1}{x_2-x_1}=m&\text{Plug in values} \\ \frac{y-y_1}{x-x_1}=m&\text{Multiply both sides by }(x-x_1) \\ y-y_1=m(x-x_1)&\text{New formula}\end{array}\nonumber\]

    Point-Slope Formula

    The point-slope formula is given by \[y-y_1=m(x-x_1),\nonumber\] given the slope \(m\) and point \((x_1, y_1)\) is on the line.

    Example \(\PageIndex{9}\)

    Using the point-slope formula, write the equation of the line passing through the point \((3, −4)\) with a slope of \(\frac{3}{5}\).

    Solution

    \[\begin{array}{rl}y-y_1=m(x-x_1)&\text{Plug values into point-slope formula} \\ y-(-4)=\frac{3}{5}(x-3)&\text{Simplify signs} \\ y+4=\frac{3}{5}(x-3)&\text{Equation in point-slope form}\end{array}\nonumber\]

    Note

    Often, we will prefer final answers be written in slope-intercept form. If the directions prefer the equations of the line in slope-intercept form, we can distribute the slope, then solve for \(y\).

    Example \(\PageIndex{10}\)

    Let’s rewrite Example \(\PageIndex{9}\) in slope-intercept form: \(y + 4 =\frac{3}{5} (x −3)\)

    Solution

    \[\begin{array}{rl}y+4=\frac{3}{5}(x-3)&\text{Distribute} \\ y+4=\frac{3}{5}x-\frac{9}{5}&\text{Isolate the variable term }y \\ y+4+\color{blue}{(-4)}\color{black}{}=\frac{3}{5}x-\frac{9}{5}+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ y=\frac{3}{5}x-\frac{11}{5}&\text{Slope-intercept form}\end{array}\nonumber\]

    Example \(\PageIndex{11}\)

    Write the equation of the line passing through the point \((−6, 2)\) with a slope of \(−\frac{2}{3}\) in slope-intercept form.

    Solution

    \[\begin{array}{rl}y-y_1=m(x-x_1)&\text{Substitute values into the point-slope formula} \\ y-2=-\frac{2}{3}(x-(-6))&\text{Simplify} \\ y-2=-\frac{2}{3}(x+6)&\text{Distribute} \\ y-2=-\frac{2}{3}x-4&\text{Isolate the variable term }y \\ y-2+\color{blue}{2}\color{black}{}=-\frac{2}{3}x-4+\color{blue}{2}\color{black}{}&\text{Simplify} \\ y=-\frac{2}{3}x-2&\text{Slope-intercept form}\end{array}\nonumber\]

    Obtaining a Line Given Two Points

    In order to find the equation of a line, we need to know the slope. If we aren’t given the slope, but only two points on the line, then we complete some preliminary work to obtain the slope. Then we can use the point-slope formula as usual to obtain the equation of the line.

    Example \(\PageIndex{12}\)

    Find the equation of the line passing through the points \((−3, 4)\) and \((−1, −2)\) in slope-intercept form

    Solution

    Since we are given two points, we can use the slope formula to obtain the slope: \[\begin{array}{rl}m=\frac{y_2-y_1}{x_2-x_1}&\text{Substitute in the ordered-pairs} \\ m=\frac{-2-4}{-1-(-3)}&\text{Simplify} \\ m=\frac{-6}{2} \\ m=-3&\text{Slope}\end{array}\nonumber\]

    Now that we have the slope, we can plug-n-chug the slope and one of the points into the pointslope formula. Notice we have two points and we can choose either one; the results will be the same. Let’s choose \((−3, 4)\) with the slope \(m = −3\).

    \[\begin{array}{rl}y-y_1=m(x-x_1)&\text{Substitute in the point and slope} \\ y-4=-3(x-(-3))&\text{Simplify} \\ y-4=-3(x+3)&\text{Distribute} \\ y-4=-3x-9&\text{Isolate the variable term }y \\ y-4+\color{blue}{4}\color{black}{}=-3x-9+\color{blue}{4}\color{black}{}&\text{Simplify} \\ y=-3x-5&\text{Slope-intercept form}\end{array}\nonumber\]

    Example \(\PageIndex{13}\)

    Find the equation of the line through the points \((6, −2)\) and \((−4, 1)\) in slope-intercept form.

    Solution

    Since we are given two points, we can use the slope formula to obtain the slope: \[\begin{array}{rl}m=\frac{y_2-y_1}{x_2-x_1}&\text{Substitute in the ordered-pairs} \\ m=\frac{1-(-2)}{-4-6}&\text{Simplify} \\ m=-\frac{3}{10}&\text{Slope}\end{array}\nonumber\]

    Now that we have the slope, we can plug-n-chug the slope and one of the points into the pointslope formula. Notice we have two points and we can choose either one; the results will be the same. Let’s choose \((−4, 1)\) with the slope \(m = −\frac{3}{10}\).

    \[\begin{array}{rl}y-y_1=m(x-x_1)&\text{Substitute in the point and slope} \\ y-3=-\frac{3}{10}(x-(-4))&\text{Simplify} \\ y-1=-\frac{3}{10}(x+4)&\text{Distribute} \\ y-1=-\frac{3}{10}x-\frac{6}{5}&\text{Isolate the variable term }y \\ y-1+\color{blue}{1}\color{black}{}=-\frac{3}{10}x-\frac{6}{5}+\color{blue}{1}\color{black}{}&\text{Simplify} \\ y=-\frac{3}{10}x-\frac{1}{5}&\text{Slope-intercept form}\end{array}\nonumber\]

    Note

    The city of Konigsberg (now Kaliningrad, Russia) had a river that flowed through the city breaking it into several parts. There were 7 bridges that connected the parts of the city. In 1735, Leonhard Euler considered the question of whether it was possible to cross each bridge exactly once and only once. It turned out that this problem was impossible, but the work laid the foundation of what would later become graph theory.

    Equations of Lines Homework

    Write the equation of the line in slope-intercept form given the slope and the \(y\)-intercept.

    Exercise \(\PageIndex{1}\)

    \(m = 2,\: y-intercept = 5\)

    Exercise \(\PageIndex{2}\)

    \(m = 1,\: y-intercept = −4\)

    Exercise \(\PageIndex{3}\)

    \(m = −\frac{3}{4},\: y-intercept = −1\)

    Exercise \(\PageIndex{4}\)

    \(m = \frac{1}{3},\: y-intercept = 1\)

    Exercise \(\PageIndex{5}\)

    \(m = −6,\: y-intercept = 4\)

    Exercise \(\PageIndex{6}\)

    \(m = −1,\: y-intercept = −2\)

    Exercise \(\PageIndex{7}\)

    \(m = −\frac{1}{4},\: y-intercept = 3\)

    Exercise \(\PageIndex{8}\)

    \(m = \frac{2}{5},\: y-intercept = 5\)

    Write the equation of the line in slope-intercept form given the graph or equation.

    Exercise \(\PageIndex{9}\)
    clipboard_e955d2e449de80481a3d9a26b85c15d76.png
    Figure \(\PageIndex{1}\)
    Exercise \(\PageIndex{10}\)
    clipboard_e7f62e92a41bf5cae7e49daf98d1397bc.png
    Figure \(\PageIndex{2}\)
    Exercise \(\PageIndex{11}\)
    clipboard_ea1e1daf8800a7e4c71be4b59e22b3733.png
    Figure \(\PageIndex{3}\)
    Exercise \(\PageIndex{12}\)
    clipboard_ed1012bbfd4c622c9cd7f920045ef0bf9.png
    Figure \(\PageIndex{4}\)
    Exercise \(\PageIndex{13}\)
    clipboard_ea390e87900e26e7568fea5bd39afd86c.png
    Figure \(\PageIndex{5}\)
    Exercise \(\PageIndex{14}\)
    clipboard_e0ba4e76dc8c4566984dbc00d50a9d4a2.png
    Figure \(\PageIndex{6}\)
    Exercise \(\PageIndex{15}\)

    \(x + 10y = −37\)

    Exercise \(\PageIndex{16}\)

    \(2x + y = −1\)

    Exercise \(\PageIndex{17}\)

    \(7x − 3y = 24\)

    Exercise \(\PageIndex{18}\)

    \(x = −8\)

    Exercise \(\PageIndex{19}\)

    \(y − 4 = −(x + 5)\)

    Exercise \(\PageIndex{20}\)

    \(y − 4 = 4(x − 1)\)

    Exercise \(\PageIndex{21}\)

    \(y + 5 = −4(x − 2)\)

    Exercise \(\PageIndex{22}\)

    \(y + 1 = −\frac{1}{2} (x − 4)\)

    Exercise \(\PageIndex{23}\)

    \(x − 10y = 3\)

    Exercise \(\PageIndex{24}\)

    \(6x − 11y = −70\)

    Exercise \(\PageIndex{25}\)

    \(4x + 7y = 28\)

    Exercise \(\PageIndex{26}\)

    \(x − 7y = −42\)

    Exercise \(\PageIndex{27}\)

    \(y − 5 = \frac{5}{2} (x − 2)\)

    Exercise \(\PageIndex{28}\)

    \(y − 3 = −\frac{2}{3} (x + 3)\)

    Exercise \(\PageIndex{29}\)

    \(0 = x − 4\)

    Exercise \(\PageIndex{30}\)

    \(y + 2 = \frac{6}{5} (x + 5)\)

    Sketch the graph of each line.

    Exercise \(\PageIndex{31}\)

    \(y = \frac{1}{3} x + 4\)

    Exercise \(\PageIndex{32}\)

    \(y = \frac{6}{5} x − 5\)

    Exercise \(\PageIndex{33}\)

    \(y = \frac{3}{2} x\)

    Exercise \(\PageIndex{34}\)

    \(x − y + 3 = 0\)

    Exercise \(\PageIndex{35}\)

    \(−y − 4 + 3x = 0\)

    Exercise \(\PageIndex{36}\)

    \(−3y = −5x + 9\)

    Exercise \(\PageIndex{37}\)

    \(y = − \frac{1}{5} x − 4\)

    Exercise \(\PageIndex{38}\)

    \(y = − \frac{3}{2} x − 1\)

    Exercise \(\PageIndex{39}\)

    \(y = − \frac{3}{4} x + 1\)

    Exercise \(\PageIndex{40}\)

    \(4x + 5 = 5y\)

    Exercise \(\PageIndex{41}\)

    \(−8 = 6x − 2y\)

    Exercise \(\PageIndex{42}\)

    \(−3y = 3 − \frac{3}{2}x\)

    Write the equation of the line in point-slope form given a point passing through the line and its slope.

    Exercise \(\PageIndex{43}\)

    \((2, 3);\: m =\text{ undefined}\)

    Exercise \(\PageIndex{44}\)

    \((2, 2);\: m = \frac{1}{2}\)

    Exercise \(\PageIndex{45}\)

    \((−1, −5);\: m = 9\)

    Exercise \(\PageIndex{46}\)

    \((−4, 1);\: m = \frac{3}{4}\)

    Exercise \(\PageIndex{47}\)

    \((0, −2);\: m = −3\)

    Exercise \(\PageIndex{48}\)

    \((0, −5);\: m = −\frac{1}{4}\)

    Exercise \(\PageIndex{49}\)

    \((−5, −3);\: m = \frac{1}{5}\)

    Exercise \(\PageIndex{50}\)

    \((−1, 4);\: m = −\frac{5}{4}\)

    Exercise \(\PageIndex{51}\)

    \((1, 2);\: m = 0\)

    Exercise \(\PageIndex{52}\)

    \((2, 1);\: m = −\frac{1}{2}\)

    Exercise \(\PageIndex{53}\)

    \((2, −2);\: m = −2\)

    Exercise \(\PageIndex{54}\)

    \((4, −3);\: m = −2\)

    Exercise \(\PageIndex{55}\)

    \((−1, 1);\: m = 4\)

    Exercise \(\PageIndex{56}\)

    \((0, 2);\: m = − \frac{5}{4}\)

    Exercise \(\PageIndex{57}\)

    \((−1, −4);\: m = − \frac{2}{3}\)

    Write the equation of the line in slope-intercept form given a point passing through the line and its slope.

    Exercise \(\PageIndex{58}\)

    \((−1, −5) ;\: m = 2\)

    Exercise \(\PageIndex{59}\)

    \((5, −1) ;\: m = − \frac{3}{5}\)

    Exercise \(\PageIndex{60}\)

    \((−4, 1) ;\: m = \frac{1}{2}\)

    Exercise \(\PageIndex{61}\)

    \((4, −2) ;\: m = − \frac{3}{2}\)

    Exercise \(\PageIndex{62}\)

    \((−5, −3) ;\: m = − \frac{2}{5}\)

    Exercise \(\PageIndex{63}\)

    \((2, −2) ;\: m = 1\)

    Exercise \(\PageIndex{64}\)

    \((−3, 4),\: m =\text{ undefined}\)

    Exercise \(\PageIndex{65}\)

    \((−4, 2) ;\: m = − \frac{1}{2}\)

    Exercise \(\PageIndex{66}\)

    \((2, −2) ;\: m = −2\)

    Exercise \(\PageIndex{67}\)

    \((−2, −2) ;\: m = − \frac{2}{3}\)

    Exercise \(\PageIndex{68}\)

    \((4, −3) ;\: m = − \frac{7}{4}\)

    Exercise \(\PageIndex{69}\)

    \((−2, 0) ;\: m = − \frac{5}{2}\)

    Exercise \(\PageIndex{70}\)

    \((3, 3) ;\: m = \frac{7}{3}\)

    Exercise \(\PageIndex{71}\)

    \((−4, −3);\: m = 0\)

    Exercise \(\PageIndex{72}\)

    \((−2, −5) ;\: m = 2\)

    Write the equation of the line in point-slope form given two points on the line.

    Exercise \(\PageIndex{73}\)

    \((−4, 3)\text{ and }(−3, 1)\)

    Exercise \(\PageIndex{74}\)

    \((5, 1)\text{ and }(−3, 0)\)

    Exercise \(\PageIndex{75}\)

    \((−4, −2)\text{ and }(0, 4)\)

    Exercise \(\PageIndex{76}\)

    \((3, 5)\text{ and }(−5, 3)\)

    Exercise \(\PageIndex{77}\)

    \((3, −3)\text{ and }(−4, 5)\)

    Exercise \(\PageIndex{78}\)

    \((1, 3)\text{ and }(−3, 3)\)

    Exercise \(\PageIndex{79}\)

    \((−4, 5)\text{ and }(4, 4)\)

    Exercise \(\PageIndex{80}\)

    \((−4, 1)\text{ and }(4, 4)\)

    Exercise \(\PageIndex{81}\)

    \((−1, −4)\text{ and }(−5, 0)\)

    Write the equation of the line in slope-intercept form given two points on the line.

    Exercise \(\PageIndex{82}\)

    \((−5, 1)\text{ and }(−1, −2)\)

    Exercise \(\PageIndex{83}\)

    \((−5, 5)\text{ and }(2, −3)\)

    Exercise \(\PageIndex{84}\)

    \((4, 1)\text{ and }(1, 4)\)

    Exercise \(\PageIndex{85}\)

    \((0, 2)\text{ and }(5, −3)\)

    Exercise \(\PageIndex{86}\)

    \((0, 3)\text{ and }(−1, −1)\)

    Exercise \(\PageIndex{87}\)

    \((−5, −1)\text{ and }(5, −2)\)

    Exercise \(\PageIndex{88}\)

    \((1, −1)\text{ and }(−5, −4)\)

    Exercise \(\PageIndex{89}\)

    \((0, 1)\text{ and }(−3, 0)\)

    Exercise \(\PageIndex{90}\)

    \((0, 2)\text{ and }(2, 4)\)


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