Graphing Vertical Lines
To graph a vertical line in the standard coordinate system, use the equation \( x = k ,\) where \(k\) gives the point on the \(x\)-axis that the line will intersect.
Why This Occurs
Suppose we want a vertical line that passes through \( (5,0) .\) In addition to that point, the line will pass through
\[ \ldots, (5, -3), (5, -2), (5, -1), (5,0), (5, 1), (5, 2), (5, 3), \ldots \]
and in general for any real number \(Q,\) the graph will pass through \( (5,Q) .\)
This means the \(y\)-coordinate can vary to any real number, so it doesn't get fixed at all and doesn't need to appear in the vertical line equation. \(x\) on the other hand must always be 5, giving an equation of \( x = 5 .\)
The Slope of a Vertical Line
Note that if we attempt to use a more traditional format, like the slope-intercept form \( y = mx + b ,\) we won't be able to form a vertical line. While the slope of \(m\) could be made very large so a graphing utility makes a line look vertical, an actual vertical line has a slope that is undefined. If we think in terms of \( \frac{\text{rise}}{\text{run}} ,\) the "run" of a vertical line must be zero; but this would cause division by zero, which is an impossible operation.
Note this is different from the slope of a horizontal line, which is 0. This would indicate a "rise" of 0. 0 in the numerator is fine: 0 divided by a real number (other than 0) is still 0. It's only division by zero that is prohibited.
Suppose instead of graphing a line as \( y = mx + b \) (where \(m\) is the slope and \(b\) is the \(y\)-intercept) we graphed it as \( x = Qy + Z .\) (Incidentally, this allows graphing vertical lines in the format.)
What is the relationship between \(m\) and \(Q?\)