The equations of three lines are given below. Line 1: 8x+6y=2 Line 2: 3y=4x+5 Line 3: y=(3)/(4)x-6 For each pair of lines, determine whether they are parallel, perpendicular, or nelther.

## Step 1: Find the slope of Line 1<br />### Rewrite $8x + 6y = 2$ in slope-intercept form ($y = mx + b$). Subtract $8x$ from both sides: $6y = -8x + 2$. Divide both sides by 6: $y = -\frac{8}{6}x + \frac{2}{6}$. Simplify: $y = -\frac{4}{3}x + \frac{1}{3}$. The slope of Line 1 is $m_1 = -\frac{4}{3}$.<br /><br />## Step 2: Find the slope of Line 2<br />### Rewrite $3y = 4x + 5$ in slope-intercept form. Divide both sides by 3: $y = \frac{4}{3}x + \frac{5}{3}$. The slope of Line 2 is $m_2 = \frac{4}{3}$.<br /><br />## Step 3: Find the slope of Line 3<br />### The equation $y = \frac{3}{4}x - 6$ is already in slope-intercept form. The slope of Line 3 is $m_3 = \frac{3}{4}$.<br /><br />## Step 4: Compare Line 1 and Line 2<br />### The slopes are $m_1 = -\frac{4}{3}$ and $m_2 = \frac{4}{3}$. Since $m_1 \neq m_2$, the lines are not parallel. Since $m_1 \times m_2 = -\frac{4}{3} \times \frac{4}{3} = -\frac{16}{9} \neq -1$, the lines are not perpendicular. Therefore, Line 1 and Line 2 are neither parallel nor perpendicular.<br /><br />## Step 5: Compare Line 1 and Line 3<br />### The slopes are $m_1 = -\frac{4}{3}$ and $m_3 = \frac{3}{4}$. Since $m_1 \neq m_3$, the lines are not parallel. Since $m_1 \times m_3 = -\frac{4}{3} \times \frac{3}{4} = -1$, the lines are perpendicular.<br /><br />## Step 6: Compare Line 2 and Line 3<br />### The slopes are $m_2 = \frac{4}{3}$ and $m_3 = \frac{3}{4}$. Since $m_2 \neq m_3$, the lines are not parallel. Since $m_2 \times m_3 = \frac{4}{3} \times \frac{3}{4} = 1 \neq -1$, the lines are not perpendicular. Therefore, Line 2 and Line 3 are neither parallel nor perpendicular.