Question

The equations of three lines are given below.Line 1: y=3/4x+7Line 2: 4y = 3x+7Line 3: 8x + 6y = -2For each pair of lines, determine whether they are parallel, perpendicular, or neither.Line1 and Line 2:Line1 and Line 3:Line2 and Line 3:

Answer 1

To answer this question, we first need to remember that:

• Two lines are parallel if and only if their slopes are equal, that is:

$m_1=m_2$

Two lines are perpendicular if and only if their slopes fulfil:

$m_1m_2=-1$

Hence, we need to find the slopes for each of the lines give. To do this we will write them in slope-intercept form:

$y=mx+b$

where m is the slope and b is the y-intercept.

Line 1.

This line is already written in slope intercept form, comparing it with the equation above we conclude that the slope of line 1 is 3/4, that is:

$m_1=\frac{3}{4}$

Line 2.

Let's write line 2 in slope intercept form:

$\begin{gathered} 4y=3x+7 \\ y=\frac{3}{4}x+\frac{7}{4} \end{gathered}$

Comparing the last line with the slope-intercept equation we conclude that line 2 has slope 3/4, that is:

$m_2=\frac{3}{4}$

Line 3.

Let's write line 3 in the appropriate form:

$\begin{gathered} 8x+6y=-2 \\ 6y=-8x-2 \\ y=-\frac{8}{6}x-\frac{2}{6} \\ y=-\frac{4}{3}x-\frac{1}{3} \end{gathered}$

From the last equivalent equation, we conclude that the slope of line 3 is equal to -4/3, that is:

$m_3=-\frac{4}{3}$

Now, that we know each slope we can determine which lines are parallel, perpendicular or neither.

Line 1 and Line 2.

We notice that the slopes of lines 1 and 2 are equal since both of them are 3/4, this means that these lines are parallel.

Line 1 and Line 3.

Since the slope are different, we will check if they are perpendicular, let's use the condition stated above:

$\begin{gathered} m_1m_3=-1 \\ (\frac{3}{4})(-\frac{4}{3})=-1 \\ -\frac{12}{12}=-1 \\ -1=-1 \end{gathered}$

Since, the condition is fulfilled we conclude that lines 1 and 3 are perpendicular.

Line 2 and Line 3.

It is clear that these slopes are not equal, let's check for if they are perpendicular:

$\begin{gathered} m_2m_3=-1 \\ (\frac{3}{4})(-\frac{4}{3})=-1 \\ -\frac{12}{12}=-1 \\ -1=-1 \end{gathered}$

Hence, lines 2 and 3 are perpendicular.

Note: Since lines 1 and 2 are parallel and lines 1 and 3 are perpendicular this readily implies that lines 2 and 3 are perpendicular as well.

Summing up:

Line 1 and Line 2: Parallel

Line 1 and Line 3: Perpendicular

Line 2 and Line 3: Perpendicular