Question

the equations of three lines are given below.
line 1: $3y = 2x + 7$
line 2: $y = \frac{2}{3}x - 4$
line 3: $6x - 4y = -4$
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: Convert to slope - intercept form

For a line in the form \(y = mx + b\), \(m\) is the slope.

• For Line 1: \(3y=2x + 7\), divide both sides by 3, we get \(y=\frac{2}{3}x+\frac{7}{3}\). So the slope of Line 1, \(m_1=\frac{2}{3}\).
• Line 2 is already in slope - intercept form \(y = \frac{2}{3}x-4\), so the slope of Line 2, \(m_2=\frac{2}{3}\).
• For Line 3: \(6x - 4y=-4\), first, we rewrite it as \(- 4y=-6x - 4\), then divide both sides by \(-4\), we get \(y=\frac{6}{4}x + 1=\frac{3}{2}x+1\). So the slope of Line 3, \(m_3=\frac{3}{2}\).

Step2: Analyze Line 1 and Line 2

Two lines are parallel if their slopes are equal (\(m_1 = m_2\)) and perpendicular if the product of their slopes is \(- 1\) (\(m_1\times m_2=-1\)).
Since \(m_1=\frac{2}{3}\) and \(m_2=\frac{2}{3}\), \(m_1 = m_2\). So Line 1 and Line 2 are parallel.

Step3: Analyze Line 1 and Line 3

Calculate the product of their slopes: \(m_1\times m_3=\frac{2}{3}\times\frac{3}{2}=1
eq - 1\), and \(m_1=\frac{2}{3}
eq m_3=\frac{3}{2}\). So Line 1 and Line 3 are neither parallel nor perpendicular.

Step4: Analyze Line 2 and Line 3

Calculate the product of their slopes: \(m_2\times m_3=\frac{2}{3}\times\frac{3}{2}=1
eq - 1\), and \(m_2=\frac{2}{3}
eq m_3=\frac{3}{2}\). So Line 2 and Line 3 are neither parallel nor perpendicular.

Answer:

Line 1 and Line 2: Parallel
Line 1 and Line 3: Neither
Line 2 and Line 3: Neither