66e1644c-b87c-47e5-8393-734297864413.html

<h4>Activity (20 minutes)</h4>
<p>Students will use knowledge of point-slope form to find the equation of lines given a point and a slope. They will then write each equation in slope-intercept form and explore equations with the same slopes and different \(y\)-intercepts. Being exposed to examples and non-examples will help lead them to the concept of perpendicular lines that have slopes that are opposite reciprocals of each other.</p>
<h4>Launch</h4>
<p>Allow students to work in pairs to go through each step of the activity. Each student should receive a paper of coordinate grids. Colored pencils and a straight-edge are recommended. When graphing, it is suggested that students graph each line in a different color and label the line with the equation. At the end of the activity, have pairs share their results of each step to the class and what they noticed along the way.</p>
<h4>Student Activity</h4>
<ol class="os-raise-noindent">
  <li>Find the equation of the line using point-slope form: \(m = 2\), \((3,1)\)</li>
</ol>
<p><strong>Answer:</strong> \(y − 1 = 2(x − 3)\)</p>
<ol class="os-raise-noindent" start="2">
  <li>Rewrite your equation from Question 1 in slope-intercept form.</li>
</ol>
<p><strong>Answer:</strong> \(y = 2x − 5\)</p>
<ol class="os-raise-noindent" start="3">
  <li>Find the equation of the line using point-slope form: \(m = -\frac12\), \((−1,4)\)</li>
</ol>
<p><strong>Answer:</strong> \(y − 4 = -\frac{1}{2}(x + 1)\)</p>
<ol class="os-raise-noindent" start="4">
  <li>Rewrite your equation from Question 3 in slope-intercept form.</li>
</ol>
<p><strong>Answer:</strong> \(y = -\frac{1}{2}x + \frac72\)</p>
<ol class="os-raise-noindent" start="5">
  <li> Find the equation of the line using point-slope form: \(m = −4\), \((2,1)\)</li>
</ol>
<p><strong>Answer:</strong> \(y − 1 = −4(x − 2)\)</p>
<ol class="os-raise-noindent" start="6">
  <li>Rewrite your equation from Question 5 in slope-intercept form.</li>
</ol>
<p><strong>Answer:</strong> \(y = −4x + 9\)</p>
<ol class="os-raise-noindent" start="7">
  <li>Find the equation of the line using point-slope form: \(m = \frac14\), \((3,2)\)</li>
</ol>
<p><strong>Answer:</strong> \(y − 2 = \frac{1}{4}(x − 3)\)</p>
<ol class="os-raise-noindent" start="8">
  <li>Rewrite your equation from Question 7 in slope-intercept form.</li>
</ol>
<p><strong>Answer:</strong> \(y = \frac{1}{4}x + \frac{5}{4}\)</p>
<ol class="os-raise-noindent" start="9">
  <li>Use the graphing tool or technology outside the course. Graph your equations from questions 2 and 4 on the same graph using the Desmos tool below. </li>
</ol>
<p>(Students were provided access to Desmos.)</p>
<p><strong>Answer:</strong>&nbsp;</p>
<p><img src="https://k12.openstax.org/contents/raise/resources/cb1e93f9695b712fb731b9b4d6c497611966ea47" alt="Graphs of y = 2x-5 and y =-(½)x+72." width="300"/></p>
<ol class="os-raise-noindent" start="10">
  <li>Use the graphing tool or technology outside the course. Graph your equations from questions 6 and 8 using the Desmos tool below. </li>
</ol>
<p>(Students were provided access to Desmos.)</p>
<p><strong>Answer:</strong>&nbsp;</p>
<p><img src="https://k12.openstax.org/contents/raise/resources/469d81b9069bf0e50fdf3432fa337ed0555daa4b" alt="Graphs of y = -4x+9 and y = (¼)x+(5/4)." width="300"/></p>
<br>
<br>
<div class="os-raise-graybox">
  <p>Perpendicular lines have slopes that have opposite signs and are reciprocals of each other. Here are some examples and non-examples of the slopes of lines that are perpendicular:</p>
  <table class="os-raise-midsizetable">
    <caption>
    Slopes of Perpendicular Lines
    </caption>
    <thead>
      <tr>
        <th scope="col">Examples</th>
        <th scope="col">Non-Examples</th>
      </tr>
    </thead>
    <tbody>
      <tr>
        <td>\( 5, -\frac{1}{5}\)</td>
        <td>\( 5, -5 \)</td>
      </tr>
      <tr>
        <td>\( -\frac{1}{7}, 7 \)</td>
        <td>\( -7, -\frac{1}{7} \)</td>
      </tr>
      <tr>
        <td>\( \frac{4}{9}, -\frac{9}{4} \)</td>
        <td>\( \frac{4}{9}, -\frac{4}{9} \)</td>
      </tr>
    </tbody>
  </table>
  <br>
</div>
<br>
<ol class="os-raise-noindent" start="11">
  <li>What is the slope of the line perpendicular to a line with a slope of \(\frac14\)?</li>
</ol>
<p><strong>Answer:</strong> The slope of the perpendicular line is -4.&nbsp;</p>
<ol class="os-raise-noindent" start="12">
  <li>What is the slope of the line perpendicular to a line with a slope of -8?</li>
</ol>
<p><strong>Answer:</strong> The slope of the perpendicular line is \(\frac18\).</p>
<ol class="os-raise-noindent" start="13">
  <li>What is the slope of the line perpendicular to a line with a slope of \(-\frac{6}{11}\)?</li>
</ol>
<p><strong>Answer:</strong> The slope of the perpendicular line is \(\frac{11}{6}\). </p>
<br>
<br>
<h4>Activity Synthesis</h4>
<p>Ask students to write an equation of a line that has a slope of zero (students can choose their \(y\)-intercept).</p>
<ul>
  <li>\(y = 0x+b\) becomes \(y=b\).</li>
  <li>What is an equation of a line parallel to \(y=b\)? Can you name a few?</li>
  <li>What would a line perpendicular to \(y=b\) look like?</li>
</ul>
<p>This will start to introduce the next concept of finding equations of lines parallel and perpendicular to the axes.</p>
<h4>1.14.4: Self Check </h4>
<p class="os-raise-text-bold"><em>After the activity, students will answer the following question to check their understanding of the concepts explored in the activity.</em></p>
<p class="os-raise-text-bold">QUESTION:</p>
<p>Which equation represents an equation of a line perpendicular to the line \( y = \frac{1}{2} x - 3 \) that contains the point \( (6, 4) \)?</p>
<table class="os-raise-textheavytable">
  <thead>
    <tr>
      <th scope="col">Answers</th>
      <th scope="col">Feedback</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td>\( y = -2x + 16  \)</td>
      <td><p>That&rsquo;s correct! Check yourself: </p>
        <p>The slope of the perpendicular line is the opposite reciprocal of the slope of the original line, \( -2 \). </p>
        <p>Substitute \( -2 \) for \( m \), 6 for \( x_1 \), and 4 for \( y_1 \) in the point-slope form: \( y - y_1 = m(x - x_1) \). </p>
        <p>Simplify the equation \( y - 4 = -2 (x - 6) \) to get \( y - 4 = -2x + 12 \).</p></td>
    </tr>
    <tr>
      <td>\( y = -2x + 14 \)</td>
      <td> Incorrect. Let&rsquo;s try again a different way:<br>
        <br>
        The slope of the perpendicular line is the opposite reciprocal of the slope of the original line, \( -2 \). The point on the line is \( (6, 4) \). The first coordinate, 6, is the \( x \)-coordinate. The second coordinate, 4, is the \( y \)-coordinate.<br>
        <br>
        Substitute \( -2 \) for \( m \), 6 for \( x_1 \), and 4 for \( y_1 \) in the point-slope form: \( y - y_1 = m(x - x_1) \).<br>
        <br>
        Simplify the equation \( y - 4 = -2 (x - 6) \) to get \( y - 4 = -2x + 12 \). The answer is \( y = -2x + 16 \). </td>
    </tr>
    <tr>
      <td>\( y = 2x - 8 \)</td>
      <td> Incorrect. Let&rsquo;s try again a different way:<br>
        <br>
        The slope of the perpendicular line is the opposite reciprocal of the slope of the original line. Remember to change the sign. The original line has slope \( \frac{1}{2} \). The opposite reciprocal is \( -2 \).<br>
        <br>
        Substitute \( -2 \) for \( m \), 6 for \( x_1 \), and 4 for \( y_1 \) in the point-slope form: \( y - y_1 = m(x - x_1) \).<br>
        <br>
        Simplify the equation \( y - 4 = -2 (x - 6) \) to get \( y - 4 = -2x + 12 \). The answer is \( y = -2x + 16 \). </td>
    </tr>
    <tr>
      <td>\( y = -2x - 2  \)</td>
      <td> Incorrect. Let&rsquo;s try again a different way:<br>
        <br>
        The slope of the perpendicular line is the opposite reciprocal of the slope of the original line, \( -2 \).<br>
        <br>
        Substitute \( -2 \) for \( m \), 6 for \( x_1 \), and 4 for \( y_1 \) in the point-slope form: \( y - y_1 = m(x - x_1) \).<br>
        <br>
        Simplify the equation \( y - 4 = -2 (x - 6) \). Be sure to distribute \(-2\) to both terms inside the parentheses to get \( y - 4 = -2x + 12 \).<br>
        <br>
        The answer is \( y = -2x + 16 \). </td>
    </tr>
  </tbody>
</table>
<br>
<h4>1.14.4: Additional Resources </h4>
<p class="os-raise-text-bold"><em>The following content is available to students who would like more support based on their experience with the self check. Students will not automatically have access to this content, so you may wish to share it with those who could benefit from it.</em></p>
<h4>Finding a Perpendicular Line Given an Equation and Point</h4>
<p>In this section, you will use what you know about writing the equation of a line given the slope and a point to write the equation of perpendicular lines. The graph shows the equation of the line \( y = 2x - 3 \). The point \( P(-2, 1) \) is also plotted. You can use the fact that perpendicular lines have slopes that are opposite reciprocals to find a line that is perpendicular to line \(l\) and goes through point \( P(-2, 1) \).</p>
<p><img alt="This figure has a graph of a two straight lines on the x y-coordinate plane. The x and y-axes run from negative 8 to 8. " class="atto_image_button_text-bottom" height="450" role="presentation" src="https://k12.openstax.org/contents/raise/resources/51e2911a4ca1a2e6fa4aaaf1120031d5db87a061" width="400"> </p>
<p>From the equation, you know the slope of the line is 2. The second line will pass through \( P(-2, 1) \) and have slope \( - \frac{1}{2} \). To graph the line, start at \( (-2, 1) \) and count out the rise and run. The slope is \( \frac{rise}{run}=\frac{-1}{2} \). Count out the rise, \( -1 \), and run, 2, and plot the point. You can graph the line as shown. Line \( n \) is perpendicular to line \( l \) and goes through point \( (-2, 1) \). </p>
<p><img alt class="atto_image_button_text-bottom" height="384" role="presentation" src="https://k12.openstax.org/contents/raise/resources/00ab16d0082a04da4d03e04b06ec75e36338fc69" width="400"> </p>
<p>To find the equation of a line perpendicular to a line through a given point algebraically, you can use what you know about finding the equation of a line given the slope and a point.</p>
<p><strong>Example</strong></p>
<p>Write an equation of a line perpendicular to \( y = 2x - 3 \) that contains the point \( (-2, 1) \). Write the equation in slope-intercept form.</p>
<p><b>Step 1 - </b> Find the slope of the given line.<br>
  The line is in slope-intercept form \( y = 2x - 3 \)<br>
  \( m=2 \) </p>
<p><b>Step 2 - </b> Find the slope of the perpendicular line<br>
  perpendicular lines have the inverse slope<br>
  \( m_1 = -1/2 \) </p>
<p><b>Step 3 - </b>Identify the point<br>
  The given point is \( (-2, 1) \)<br>
  \( (x_1, y_1) = (-2, 1) \) </p>
<p><b>Step 4 - </b>Substitute values into the point-slope form \( y - y_1 = m(x - x_1) \)<br>
  Simplify<br>
  \( \begin{array}{rcl}y - y_1 &amp;=&amp; m(x - x_1) \\ y-1 &amp;=&amp; -\frac{1}{2}(x - (-2)) \\ y-1 &amp;=&amp; -\frac{1}{2}(x + 2) \\ y-1 &amp;=&amp; -\frac{1}{2}x - 1\end{array} \)</p>
<p><b>Step 5 - </b>Write the equation in slope-intercept form<br>
  \( y = -\frac{1}{2}x\) </p>
<p>Use this table for reference to write an equation of a line perpendicular to a given line: </p>
<br>
<div class="os-raise-graybox">
  <p><strong>Step 1</strong> - Find the slope of the given line.</p>
  <p><strong>Step 2</strong> - Find the slope of the perpendicular line.</p>
  <p><strong>Step 3</strong> - Identify the point.</p>
  <p><strong>Step 4</strong> - Substitute the values into the point-slope form: \( y-y_1=m(x-x_1) \).</p>
  <p><strong>Step 5</strong> - Simplify.</p>
  <p><strong>Step 6</strong> - Write the equation in slope-intercept form \( (y = mx+b) \).</p>
</div>
<br>
<br>
<h4>Try It: Finding a Perpendicular Line Given an Equation and Point</h4>
<p>Find an equation of a line perpendicular to the line \( y = 3x + 1 \) that contains the point \( (4, 2) \). Write the equation in slope-intercept form.</p>
<p>Write down your answer, and then select the<strong> solution</strong> button to compare your work.</p>
<p><strong>Answer:</strong></p>
<p>Here is how to find a line perpendicular to a given line and point.</p>
<p><strong>Step 1</strong> - Find the slope of the given line.<br>
  \( m = 3 \)</p>
<p><strong>Step 2</strong> - Find the slope of the perpendicular line. <br>
  \( m_1 = - \frac{1}{3} \)</p>
<p><strong>Step 3</strong> - Identify the point.<br>
  \( (x_1, y_1) \)<br>
  \( (4, 2) \)</p>
<p><strong>Step 4</strong> - Substitute the values into the point-slope form<br>
  \( y-y_1=m(x-x_1) \).<br>
  \(\begin{array}{rcl} y - y_1 &amp;= m(x - x_1) \\ y - 2 &amp;= -\frac{1}{3} (x - 4)\end{array} \)</p>
<p><strong>Step 5</strong> - Simplify.<br>
  \( y - 2 = -\frac{1}{3}x + \frac{4}{3} \)</p>
<p><strong>Step 6</strong> - Write the equation in slope-intercept form \( (y = mx+b) \).<br>
  \( y = -\frac{1}{3}x + 3\frac{1}{3}  \)</p>