<PRE><B>Question 189847</B>
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A line perpendicular to the &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;-axis is a vertical line.  All vertical lines have an equation of the form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = a]


meaning the set of ordered pairs where the &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;-coordinate is always &lt;i&gt;&lt;b&gt;a&lt;/b&gt;&lt;/i&gt; and the &lt;i&gt;&lt;b&gt;y&lt;/b&gt;&lt;/i&gt;-coordinate is any real number.


The &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;-intercept of the given equation is an ordered pair of the form (&lt;i&gt;&lt;b&gt;a&lt;/b&gt;&lt;/i&gt;,&lt;i&gt;&lt;b&gt;0&lt;/b&gt;&lt;/i&gt;), so you need to find the value of &lt;i&gt;&lt;b&gt;a&lt;/b&gt;&lt;/i&gt; to substitute for &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt; in *[tex \LARGE 6x-5y+18=0] so that &lt;i&gt;&lt;b&gt;y&lt;/b&gt;&lt;/i&gt; = 0.


To find this value, rearrange the equation so that you have &lt;i&gt;&lt;b&gt;y&lt;/b&gt;&lt;/i&gt; on one side of the equals sign and everything else on the other side.  Then replace &lt;i&gt;&lt;b&gt;y&lt;/b&gt;&lt;/i&gt; with 0 and solve for &lt;i&gt;&lt;b&gt;x&lt;/b&gt;&lt;/i&gt;.  This value will be your &lt;i&gt;&lt;b&gt;a&lt;/b&gt;&lt;/i&gt; in the equation above.


To me,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = a]


is Standard Form, though some may prefer


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x - a = 0]


It depends on whether Standard Form is defined as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Ax + By = C]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Ax + By + C = 0]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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