Question

the equations of three lines are given below.
line 1: 3x - 12y = 24
line 2: y = -4x + 8
line 3: y = -4x - 1
for each pair of lines, determine whether they are parallel, perpendicular, or neither.
line 1 and line 2: ∘ parallel ∘ perpendicular ∘ neither
line 1 and line 3: ∘ parallel ∘ perpendicular ∘ neither
line 2 and line 3: ∘ parallel ∘ perpendicular ∘ neither

Explanation:

Step1: Find slope of Line 1

Rewrite \(3x - 12y = 24\) in slope - intercept form \(y=mx + b\) (where \(m\) is the slope).
Subtract \(3x\) from both sides: \(- 12y=-3x + 24\).
Divide by \(-12\): \(y=\frac{-3}{-12}x+\frac{24}{-12}\), so \(y = \frac{1}{4}x-2\). The slope of Line 1, \(m_1=\frac{1}{4}\).

Step2: Analyze Line 1 and Line 2

Slope of Line 2, \(m_2=-4\).
Check if parallel: \(m_1
eq m_2\) (\(\frac{1}{4}
eq - 4\)), so not parallel.
Check if perpendicular: Two lines are perpendicular if \(m_1\times m_2=-1\). \(\frac{1}{4}\times(-4)=-1\), so Line 1 and Line 2 are perpendicular.

Step3: Analyze Line 1 and Line 3

Slope of Line 3, \(m_3 = - 4\).
Check if parallel: \(m_1
eq m_3\) (\(\frac{1}{4}
eq - 4\)), so not parallel.
Check if perpendicular: \(\frac{1}{4}\times(-4)=-1\), so Line 1 and Line 3 are perpendicular. Wait, no, wait: Wait, \(m_1=\frac{1}{4}\), \(m_3=-4\), \(\frac{1}{4}\times(-4) = - 1\), but wait, let's re - check. Wait, Line 1: \(y=\frac{1}{4}x - 2\), Line 3: \(y=-4x-1\). The product of slopes is \(\frac{1}{4}\times(-4)=-1\), but wait, no, earlier for Line 2, same slope as Line 3. Wait, no, Line 2 and Line 3 have the same slope (\(m_2 = m_3=-4\)). Let's re - do Line 1 and Line 3.
Wait, \(m_1=\frac{1}{4}\), \(m_3=-4\). \(\frac{1}{4}\times(-4)=-1\), so they are perpendicular? Wait, no, wait, maybe I made a mistake. Wait, Line 1: \(3x-12y = 24\) gives \(y=\frac{1}{4}x - 2\) (correct). Line 3: \(y=-4x-1\) (slope - 4, correct). The product of \(\frac{1}{4}\) and \(-4\) is \(-1\), so they are perpendicular? But wait, Line 2 and Line 3 have the same slope. Let's check Line 2 and Line 3.

Step4: Analyze Line 2 and Line 3

Slope of Line 2, \(m_2=-4\), slope of Line 3, \(m_3=-4\). Since \(m_2 = m_3\), Line 2 and Line 3 are parallel.

Wait, let's re - summarize:

• Line 1: \(y=\frac{1}{4}x - 2\) (slope \(\frac{1}{4}\))
• Line 2: \(y=-4x + 8\) (slope \(-4\))
• Line 3: \(y=-4x-1\) (slope \(-4\))

So:

• Line 1 and Line 2: Product of slopes \(\frac{1}{4}\times(-4)=-1\) → Perpendicular
• Line 1 and Line 3: Product of slopes \(\frac{1}{4}\times(-4)=-1\) → Perpendicular? Wait, no, wait, \(\frac{1}{4}\times(-4)=-1\), so yes, perpendicular. But wait, Line 2 and Line 3: slopes are equal (\(-4=-4\)) → Parallel.

Wait, maybe I messed up Line 1 and Line 3. Wait, let's recalculate Line 1's slope. \(3x-12y = 24\):

\(-12y=-3x + 24\) → \(y=\frac{3}{12}x-\frac{24}{12}\) → \(y=\frac{1}{4}x - 2\) (correct). So slope is \(\frac{1}{4}\). Line 2: slope \(-4\), Line 3: slope \(-4\).

So:

• Line 1 and Line 2: \(\frac{1}{4}\times(-4)=-1\) → Perpendicular
• Line 1 and Line 3: \(\frac{1}{4}\times(-4)=-1\) → Perpendicular? Wait, no, that can't be. Wait, no, if two lines have slopes that are negative reciprocals, they are perpendicular. The negative reciprocal of \(\frac{1}{4}\) is \(-4\), and vice - versa. So any line with slope \(-4\) is perpendicular to a line with slope \(\frac{1}{4}\). So Line 1 (slope \(\frac{1}{4}\)) and Line 2 (slope \(-4\)): perpendicular. Line 1 and Line 3 (slope \(-4\)): perpendicular. Line 2 and Line 3 (slope \(-4\) and \(-4\)): parallel.

Answer:

Line 1 and Line 2: Perpendicular
Line 1 and Line 3: Perpendicular
Line 2 and Line 3: Parallel

Wait, no, wait, earlier mistake: Wait, Line 1: \(y=\frac{1}{4}x-2\), Line 3: \(y = - 4x-1\). The product of slopes is \(\frac{1}{4}\times(-4)=-1\), so they are perpendicular. Line 2 and Line 3: same slope (\(-4\)), so parallel. Line 1 and Line 2: product of slopes \(-1\), so perpendicular.