Elementary Algebra

We could graph the lines, but we can't be sure from a graph if the lines are exactly parallel or exactly perpendicular. A more accurate way to answer the question is to find the slope of each line. To do this we write each equation in slope-intercept form, that is, we solve for \(y\text{.}\)

The slope of the first line is \(\dfrac{-2}{3}\text{,}\) and the slope of the second line is \(\dfrac{3}{2}\text{.}\) The slopes are not equal, so the lines are not parallel. However, the product of the slopes is

so the lines are perpendicular.

Looking at the graph of the line shown below, we see that the \(y\)-coordinate of every point on the line is 3.

In particular, the \(y\)-intercept of the line is the point \((0,3)\text{,}\) so \(b=3\text{.}\) As we noted above, the slope of the line is \(m=0\text{.}\) Substituting these values into the slope-intercept form gives us the equation

or just \(y=3\text{.}\) The fact that \(x\) does not appear in the equation means that \(y=3\) for every point on the line, no matter what the value of \(x\) is.

1. As we move from \(P\) to \(Q\text{,}\) we move up 2 squares on the graph, so \(\Delta y = 2\text{.}\) As we move from \(Q\) to \(R\text{,}\) we move 8 squares to the right, so \(\Delta x = 8\text{.}\) Thus, the slope of the line is

   \begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{2}{8} = \dfrac{1}{4} \end{equation*}

2. First we write down the coordinates of \(P\) and \(Q\text{:}\)

   \begin{equation*} P(-3,\blert{-4})~~~~~~\text{and}~~~~~~Q(-3,\blert{-2}) \end{equation*}

   and compute the directed distance from \(P\) to \(Q\text{:}\)

   \begin{equation*} \Delta y = -2-(-4) = 2~~~~~~~~~~~~~~~~~~\blert{\text{final} - \text{initial}} \end{equation*}

   Then we write down the coordinates of \(Q\) and \(R\text{:}\)

   \begin{equation*} Q(\blert{-3},-2)~~~~~~\text{and}~~~~~~R(\blert{5},-2) \end{equation*}

   and compute the directed distance from \(Q\) to \(R\text{:}\)

   \begin{equation*} \Delta x = 5-(-3) = 8~~~~~~~~~~~~~~~~~~\blert{\text{final} - \text{initial}} \end{equation*}

   We get the same value for the slope as in part (a),

   \begin{equation*} m = \dfrac{\Delta y}{\Delta x} = \dfrac{2}{8} = \dfrac{1}{4} \end{equation*}

We'll call \(D\) the first point and \(F\) the second point. Then their coordinates are

Now we can describe the formula for slope in an organized way. We compute \(\Delta y\) and \(\Delta x\) as directed distances. First, we observe that the coordinates of point \(E\) are \((-6,4)\) or \((x_1,y_2)\text{.}\) From the figure, we can see that

and

Thus, the slope of the line segment joining \(D\) and \(F\) is

It doesn't matter which point is \(P_1\) and which is \(P_2\text{,}\) so we choose \(P_1\) to be \((6,-2)\text{.}\) Then \((x_1,y_1)=(6,-2)\) and \((x_2,y_2)=(3,-1)\text{.}\) Thus,