Question

the equations of three lines are given below.
line 1: $3y = 2x + 7$
line 2: $6x - 4y = 6$
line 3: $y = \frac{2}{3}x - 5$
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 slopes of each line

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

• Line 1: \(3y = 2x + 7\) → \(y=\frac{2}{3}x+\frac{7}{3}\), so slope \(m_1=\frac{2}{3}\).
• Line 2: \(6x - 4y = 6\) → \(-4y=-6x + 6\) → \(y=\frac{6}{4}x-\frac{6}{4}=\frac{3}{2}x-\frac{3}{2}\), so slope \(m_2=\frac{3}{2}\).
• Line 3: \(y=\frac{2}{3}x - 5\), so slope \(m_3=\frac{2}{3}\).

Step2: Compare slopes for each pair

• Line 1 and Line 2: Check if \(m_1 = m_2\) (parallel) or \(m_1\times m_2=-1\) (perpendicular). \(m_1=\frac{2}{3}\), \(m_2=\frac{3}{2}\). \(\frac{2}{3}

eq\frac{3}{2}\), and \(\frac{2}{3}\times\frac{3}{2}=1
eq - 1\). So neither.

• Line 1 and Line 3: \(m_1=\frac{2}{3}\), \(m_3=\frac{2}{3}\). Since \(m_1 = m_3\), they are parallel.
• Line 2 and Line 3: \(m_2=\frac{3}{2}\), \(m_3=\frac{2}{3}\). \(\frac{3}{2}

eq\frac{2}{3}\), and \(\frac{3}{2}\times\frac{2}{3}=1
eq - 1\). So neither.

Answer:

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