4.5 Use the slope–intercept form of an equation of a line  (Page 6/12)

Solution

$\begin{array}{cccccccccc}\begin{array}{}\\ \\ \\ \\ \\ \text{Solve the first equation for}y.\hfill \\ \\ \\ \\ \\ \\ \text{The equation is now in slope–intercept form.}\hfill \\ \\ \\ \end{array}\hfill & & & \begin{array}{ccc}\hfill 3x-2y& =\hfill & 6\hfill \\ \hfill -2y& =\hfill & \mathrm{-3}x+6\hfill \\ \hfill \frac{\mathrm{-2}y}{\mathrm{-2}}& =\hfill & \frac{\mathrm{-3}x+6}{\mathrm{-2}}\hfill \\ \hfill y& =\hfill & \frac{3}{2}x-3\hfill \end{array}\hfill & & & \hfill \begin{array}{c}\hfill \text{and}\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{c}y=\frac{3}{2}x+1\hfill \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill \\ \begin{array}{c}\text{The equation of the second line is already}\hfill \\ \text{in slope–intercept form.}\hfill \\ \text{Identify the slope and}y\text{-intercept of both lines.}\hfill \\ \\ \\ \\ \\ \\ \\ \\ \end{array}\hfill & & & \begin{array}{}\\ \\ \\ \\ \\ \hfill y& =\hfill & \frac{3}{2}x-3\hfill \\ \hfill y& =\hfill & mx+b\hfill \\ \hfill m& =\hfill & \frac{3}{2}\hfill \end{array}\hfill & & & & & & \begin{array}{c}y=\frac{3}{2}x+1\hfill \\ y=\frac{3}{2}x+1\hfill \\ y=mx+b\hfill \\ m=\frac{3}{2}\hfill \end{array}\hfill \\ & & & y\text{-intercept is (0, −3)}\hfill & & & & & & y\text{-intercept is (0, 1)}\hfill \end{array}$

The lines have the same slope and different y -intercepts and so they are parallel. You may want to graph the lines to confirm whether they are parallel.

Solution

$\begin{array}{cccc}\begin{array}{}\\ \\ \\ \\ \text{Write each equation in slope–intercept form.}\hfill \\ \text{Since there is no}x\text{term we write}0x.\hfill \\ \text{Identify the slope and}y\text{-intercept of both lines.}\hfill \end{array}\hfill & & & \begin{array}{ccccccc}y=\mathrm{-4}\hfill & & & \text{and}\hfill & & & y=3\hfill \\ y=0x-4\hfill & & & & & & y=0x+3\hfill \\ \\ y=0x-4\hfill & & & & & & y=0x+3\hfill \\ y=mx+b\hfill & & & & & & y=mx+b\hfill \\ m=0\hfill & & & & & & m=0\hfill \\ y\text{-intercept is (0, 4)}\hfill & & & & & & y\text{-intercept is (0, 3)}\hfill \end{array}\hfill \end{array}$

The lines have the same slope and different y -intercepts and so they are parallel.

There is another way you can look at this example. If you recognize right away from the equations that these are horizontal lines, you know their slopes are both 0. Since the horizontal lines cross the y -axis at $y=\mathrm{-4}$ and at $y=3$ , we know the y -intercepts are $(0,\mathrm{-4})$ and $\left(0,3\right)$ . The lines have the same slope and different y -intercepts and so they are parallel.

Solution

Since there is no $y$ , the equations cannot be put in slope–intercept form. But we recognize them as equations of vertical lines. Their x -intercepts are $\mathrm{-2}$ and $\mathrm{-5}$ . Since their x -intercepts are different, the vertical lines are parallel.