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      <text>12.1.1.1 Introduction and Perpendicular lines</text>
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      <text><![CDATA[<h3>Perpendicular lines</h3>
<p>One form for the equation of a straight line is</p>
<p>\(\qquad y=mx+c\)</p>
<p>where \(m\) and \(c\) are constants. We remember that \(m\) is the gradient of the line and its value is the tangent of the angle \(\theta\) that the line makes with the positive \(x\)-axis. The constant \(c\) is the value obtained where the line intersects
    the \(y\)-axis. See figure below:</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-1-fig-1-PerpLine.svg" alt="" width="300" height="152" class="img-responsive atto_image_button_text-bottom"></figure>
<p style="text-align: left;">If we have a second line, with equation <br></p>
<p style="text-align: left;">\(\qquad y=nx+d\)</p>
<p style="text-align: left;"> then, unless \(m=n\), the two lines will intersect at one point. These are drawn together in the figure below. The second line makes an angle \(\psi\) with the positive \(x\)-axis.</p>
<figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-1-fig-2-PerpLine.svg" alt="" width="300" height="179" class="img-responsive atto_image_button_text-bottom"></figure>
  <p>
  A simple question to ask is "what is the relation between \(m\) and \(n\) if the lines are perpendicular?" If the lines are perpendicular, as shown in the figure below, the angles \(\theta\) and \(\psi\) must satisfy the relation:</p>
    <p>\(\qquad \psi -\theta=90^{\circ}\)</p>
    <figure class="HELM_figure"><img src="@@PLUGINFILE@@/12-1-fig-3-PerpLine.svg" alt="" width="300" height="179" class="img-responsive atto_image_button_text-bottom"></figure>
    <p>This is true since the angles in a triangle add up to \(180^{\circ}\). According to the figure, the three angles are \(90^{\circ}\), \(\theta\) and \(180^{\circ}-\psi\). Therefore</p>
    <p> \(\qquad 180^{\circ}= 90^{\circ}+\theta +(180^{\circ}-\psi)\qquad \text{implying}\qquad \psi -\theta =90^{\circ}\)<br></p>
    <p>In this special case that the lines are perpendicular or <strong>normal</strong> to each other, the relation between the gradients \(m\) and \(n\) is easily obtained. In this deduction we use the following basic trigonometric relations and identities:&nbsp;</p>
    <p>\(\qquad\sin(A-B)\equiv\sin A\cos B-\cos A\sin B\qquad \cos (A-B)\equiv\cos A\cos B +\sin A\sin B\)</p>
    <p>\(\qquad\tan(A)\equiv\dfrac{\sin A}{\cos A}\qquad \sin 90^{\circ}=1 \qquad \cos 90^{\circ}=0\)&nbsp;</p>
    <p><span>


Now</span> \(\begin{aligned} m&amp;=\tan\theta\\ &amp;=\tan(\psi-90^o)\qquad\text{(see Figure)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &amp;=\frac{\sin(\psi-90^o)}{\cos(\psi-90^o)}\\ &amp;=-\frac{\cos\psi}{\sin\psi}=-\frac{1}{\tan\psi}=-\frac{1}{n}\\
        \text{So}\qquad mn&amp;=-1\end{aligned}\)</p>

    <div class="HELM_keypoint">
        <h4>Key Point</h4>
        <p>Two straight lines \(y = mx + c,\ y = nx + d\) are perpendicular if </p>
        <p style="text-align: center;">
            \(m = -\frac{1}{n} \) or equivalently \(mn = -1\)</p>
        <p style="text-align: left;">This result assumes that neither of the lines are parallel to the \(x\)-axis or to the \(y\)-axis, as in such cases one gradient will be zero and the other infinite.</p>
    </div>]]></text>
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