Question

the equations of three lines are given below.
line 1: $8x + 6y = -8$
line 2: $y = \frac{3}{4}x + 2$
line 3: $3y = 4x + 7$
for each pair of lines, determine whether they are parallel, perpendicular, or neither.
line 1 and line 2: $\circ$ parallel $\circ$ perpendicular $\circ$ neither
line 1 and line 3: $\circ$ parallel $\circ$ perpendicular $\circ$ neither
line 2 and line 3: $\circ$ parallel $\circ$ perpendicular $\circ$ neither

Explanation:

Step1: Find slope of Line 1

Rewrite \(8x + 6y=-8\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
Subtract \(8x\) from both sides: \(6y=-8x - 8\).
Divide by 6: \(y=-\frac{8}{6}x-\frac{8}{6}=-\frac{4}{3}x-\frac{4}{3}\). So slope of Line 1 (\(m_1\)) is \(-\frac{4}{3}\).

Step2: Identify slope of Line 2

Line 2 is \(y = \frac{3}{4}x + 2\), so slope of Line 2 (\(m_2\)) is \(\frac{3}{4}\).

Step3: Find slope of Line 3

Rewrite \(3y = 4x+7\) in slope - intercept form. Divide by 3: \(y=\frac{4}{3}x+\frac{7}{3}\). So slope of Line 3 (\(m_3\)) is \(\frac{4}{3}\).

Analyze Line 1 and Line 2:

Check if \(m_1\times m_2=- 1\) (perpendicular) or \(m_1 = m_2\) (parallel).
\(m_1\times m_2=(-\frac{4}{3})\times\frac{3}{4}=- 1\). So Line 1 and Line 2 are perpendicular.

Analyze Line 1 and Line 3:

\(m_1=-\frac{4}{3}\), \(m_3=\frac{4}{3}\). \(m_1
eq m_3\) (not parallel) and \(m_1\times m_3=(-\frac{4}{3})\times\frac{4}{3}=-\frac{16}{9}
eq - 1\) (not perpendicular). So Line 1 and Line 3 are neither.

Analyze Line 2 and Line 3:

\(m_2=\frac{3}{4}\), \(m_3=\frac{4}{3}\). \(m_2
eq m_3\) (not parallel) and \(m_2\times m_3=\frac{3}{4}\times\frac{4}{3}=1
eq - 1\) (not perpendicular). So Line 2 and Line 3 are neither.

Answer:

Line 1 and Line 2: Perpendicular
Line 1 and Line 3: Neither
Line 2 and Line 3: Neither