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Among the family of linear relationships, there are some subgroups of lines that are noteworthy. Two such types are horizontal and vertical lines.

A horizontal line is a line that runs parallel to the $x-$axis. An example can be seen in the diagram below.

Every point on the line above has $y-$coordinate $2.$ In fact, the rule for this function is $y=2.$ Notice that this looks different than most linear functions. The slope $m$ of a line is the quotient between the rise and the run between any two of its points. $m=runrise $ In the graph above, it can be seen that the line does not rise. In other words, the vertical change between any two points is $0.$ Therefore, its slope is $0.$ In fact, all horizontal lines have slope $m=0.$ The slope-intercept form of horizontal lines can be written as follows. $y=0⋅x+b⇔y=b $

Therefore, all horizontal lines can be written in this form, where $b$ is the $y-$intercept.A vertical line is a line that runs parallel to the $y-$axis. A vertical line is drawn in the coordinate plane below.

Notice that the line passes through all points where the $x-$coordinate is $3.$ This means that the same $x-$value corresponds to infinitely many $y-$values. Therefore, this line does not represent a function. In fact, no vertical line is a function. Similar to horizontal lines, vertical lines take the following form.

$x=a$

One horizontal line and one vertical line intersect at the point $(1,-6).$ Write the equations of the lines.

Show Solution

To write the equations of the horizontal and vertical lines that intersect at $(1,-6),$ it can be helpful to first sketch a graph. We'll plot the point, then draw a vertical and horizontal line through the point.

The horizontal line intersects the $y$-axis at $-6$. Thus, we can write the rule as $y=-6.$ The vertical line intersects the $x$-axis at $1.$ Thus, its equation is $x=1.$ Therefore, the horizontal and vertical lines that intersect at $(1,-6)$ are $y=-6andx=1.$

Together with the axes, a horizontal and a vertical line form a square in Quadrant I of the coordinate plane. Write the rules for the lines so that the square has an area of $16$ square units.

Show Solution

To begin, let's consider the area of a square. One formula that can be used to calculate the area of a square is $A=s_{2},$ where $s$ is the side length of the square. The square's area, $A,$ is $16$ square units. Therefore, we must find a side that when squared equals $16.$ We recognize that $16$ is a perfect square and that $16=4_{2}.$ Therefore, each side of the square must measure $4$ units. Let's sketch a horizontal and vertical line that, together with the axes, for a square with side length 4. Notice that the $x$-axis and $y$-axis will be the bottom and left side of the square, respectively. To ensure that the top and bottom of the square are $4$ units, we can draw a vertical line at $x=4.$

Next, to ensure that the sides are also $4$ units, we can draw a horizontal line at $y=4.$

The horizontal line intercepts the $y$-axis at $4.$ Therefore, the line can be written as $y=4.$ The vertical line intercepts the $x$-axis at $4.$ Thus, we can write the vertical line as $x=4.$ The two lines that form the requested square are
$y=4andx=4.$

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