Elementary Algebra

1. We plot the point \((1,3)\text{,}\) then use the slope, \(-2\text{,}\) to find another point on the line. From the point \((1,3)\text{,}\) we move 2 units down and 1 unit to the right, arriving at \((2,1)\text{.}\) We draw the line through these two points. The graph is shown below.

   

2. We use the formula

   \begin{equation*} \dfrac{y-y_1}{x-x_1} = m \end{equation*}

   with \(\dfrac{\Delta y}{\Delta x}=\dfrac{-2}{1}\) and \((x_1,y_1)=(\alert{1},\alert{3})\) to get

   \begin{equation*} \dfrac{-2}{1}=\dfrac{y-\alert{3}}{x-\alert{1}} \end{equation*}

   To simplify the equation, we cross-multiply.

   \begin{align*} 1(y-3) \amp = -2(x-1) \amp\amp \blert{\text{Apply the distributive law.}}\\ y-3\amp = -2x+2 \amp\amp \blert{\text{Add 3 to both sides.}}\\ y \amp = -2x+5 \end{align*}

   You can verify on the graph that \(y=-2x+5\) is an equation for the line.

We know one point on the required line, but we also need to know its slope. Now, our line is perpendicular to the given line, and the slope of the given line is \(~m_1=2~\) (as you can check by putting its equation into slope-intercept form). So the slope of our line is

Now we can use the point-slope formula with \(~m_1=\dfrac{-1}{2}\) and \((x_1,y_1) = (1,4)\text{.}\)

The given line and the perpendicular line are shown in the figure.

We solve this problem in two steps: First, we find the slope of the line, and then we use the point-slope formula.

Step 1 Let \((x_1,y_1) = (2,-1)\) and \((x_2,y_2) = (-1,3)\text{.}\) Using the slope formula, we find

Step 2 We apply the point-slope formula with \(m=\dfrac{-4}{3}\) and \((x_1,y_1) = (2,-1)\text{.}\) (We can use either point to find the equation of the line.) Then

We cross-multiply to find

The graph of the line is shown in the figure.

1. We have two data points of the form \((t,W)\text{,}\) namely \((t_1,W_1)=(3,188.6)\) and \((t_2,W_2)=(13,258.3)\text{.}\) We use the point slope formula to fit a line through these two points. First we compute the slope.

   \begin{equation*} \dfrac{\Delta W}{\Delta t}=\dfrac{W_2-W_2}{t+2-t_1}=\dfrac{258.3-188.6}{13-3}=6.97 \end{equation*}

   Next, we substitute the slope \(m=6.97\) and either of the two data points into the point-slope formula.

   \begin{align*} W \amp = W_1+m(t-t_1)\\ W \amp = 188.6+6.97(t-3)\\ W \amp = 167.69+6.97t \end{align*}

   Thus, \(W = 167.69+6.97t\text{.}\)

2. The slope gives us the rate of change of \(W\) with respect to \(t\text{.}\) The units of the variables can help us interpret the slope in context.

   \begin{equation*} \dfrac{\Delta W}{\Delta t}=\dfrac{258.3-188.6~\text{millions of cases}}{13-3~\text{years}} = 6.97~ \text{millions of cases/years} \end{equation*}

   Over the ten years between 1993 and 2003, wine consumption in the US increased at a rate of 6.97 million cases per year.