Question

the equations of three lines are given below. line 1: $4x - 10y = -6$ line 2: $2y = -5x + 3$ line 3: $y = -\frac{5}{2}x - 8$ 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 \(4x - 10y=-6\) in slope - intercept form \(y = mx + b\) (where \(m\) is the slope).
\(-10y=-4x - 6\)
\(y=\frac{-4}{-10}x+\frac{-6}{-10}\)
\(y=\frac{2}{5}x+\frac{3}{5}\), so slope of Line 1 (\(m_1\)) is \(\frac{2}{5}\).

Step2: Find slope of Line 2

Rewrite \(2y=-5x + 3\) in slope - intercept form.
\(y=\frac{-5}{2}x+\frac{3}{2}\), so slope of Line 2 (\(m_2\)) is \(-\frac{5}{2}\).

Step3: Find slope of Line 3

Line 3 is \(y =-\frac{5}{2}x-8\), so slope of Line 3 (\(m_3\)) is \(-\frac{5}{2}\).

Step4: Analyze Line 1 and Line 2

Check if \(m_1\times m_2=- 1\) (perpendicular) or \(m_1 = m_2\) (parallel).
\(m_1\times m_2=\frac{2}{5}\times(-\frac{5}{2})=-1\), so Line 1 and Line 2 are perpendicular.

Step5: Analyze Line 1 and Line 3

Check \(m_1\) and \(m_3\). \(\frac{2}{5} eq-\frac{5}{2}\) and \(\frac{2}{5}\times(-\frac{5}{2})=-1\)? Wait, no, \(\frac{2}{5}\times(-\frac{5}{2})=-1\) but \(m_1=\frac{2}{5}\), \(m_3 =-\frac{5}{2}\), wait no, in step 4 we saw Line 1 and Line 2 are perpendicular. For Line 1 and Line 3: \(\frac{2}{5} eq-\frac{5}{2}\) and \(\frac{2}{5}\times(-\frac{5}{2})=-1\)? Wait, no, \(\frac{2}{5}\times(-\frac{5}{2})=-1\), but wait, no, Line 1 slope is \(\frac{2}{5}\), Line 3 slope is \(-\frac{5}{2}\). Wait, \(\frac{2}{5}\times(-\frac{5}{2})=-1\), but wait, no, in step 2 Line 2 slope is \(-\frac{5}{2}\), Line 3 slope is also \(-\frac{5}{2}\). Wait, I made a mistake. Wait Line 1: \(y=\frac{2}{5}x+\frac{3}{5}\), slope \(m_1=\frac{2}{5}\). Line 2: \(y =-\frac{5}{2}x+\frac{3}{2}\), slope \(m_2=-\frac{5}{2}\). Line 3: \(y=-\frac{5}{2}x - 8\), slope \(m_3=-\frac{5}{2}\).

So for Line 1 and Line 2: \(m_1\times m_2=\frac{2}{5}\times(-\frac{5}{2})=-1\), so perpendicular.

Line 1 and Line 3: \(m_1=\frac{2}{5}\), \(m_3 =-\frac{5}{2}\). \(\frac{2}{5} eq-\frac{5}{2}\) and \(\frac{2}{5}\times(-\frac{5}{2})=-1\)? Wait, no, \(\frac{2}{5}\times(-\frac{5}{2})=-1\), but wait, no, if two lines are perpendicular, their slopes multiply to - 1. But wait, Line 1 slope is \(\frac{2}{5}\), Line 3 slope is \(-\frac{5}{2}\), \(\frac{2}{5}\times(-\frac{5}{2})=-1\)? Wait, yes, but wait Line 2 and Line 3 have the same slope. Wait, no, I think I messed up. Wait Line 1: \(y=\frac{2}{5}x + 0.6\), Line 3: \(y=-\frac{5}{2}x-8\). \(\frac{2}{5}\times(-\frac{5}{2})=-1\), but wait, no, \(\frac{2}{5}\) and \(-\frac{5}{2}\) are negative reciprocals. But wait, Line 2 and Line 3 have the same slope (\(-\frac{5}{2}\)).

Wait, let's re - check:

Line 1: \(4x-10y=-6\)

\(-10y=-4x - 6\)

\(y=\frac{4}{10}x+\frac{6}{10}=\frac{2}{5}x+\frac{3}{5}\). Correct, slope \(\frac{2}{5}\).

Line 2: \(2y=-5x + 3\)

\(y=-\frac{5}{2}x+\frac{3}{2}\). Slope \(-\frac{5}{2}\). Correct.

Line 3: \(y =-\frac{5}{2}x-8\). Slope \(-\frac{5}{2}\). Correct.

So Line 1 and Line 2: slopes multiply to \(\frac{2}{5}\times(-\frac{5}{2})=-1\), so perpendicular.

Line 1 and Line 3: \(\frac{2}{5} eq-\frac{5}{2}\) and \(\frac{2}{5}\times(-\frac{5}{2})=-1\)? Wait, yes, but wait, no, \(\frac{2}{5}\) and \(-\frac{5}{2}\) are negative reciprocals, so their product is - 1, but wait, Line 3's slope is \(-\frac{5}{2}\), same as Line 2. Wait, no, I think I made a mistake in the product. \(\frac{2}{5}\times(-\frac{5}{2})=-1\), so Line 1 and Line 3: are they perpendicular? Wait, Line 1 slope \(\frac{2}{5}\), Line 3 slope \(-\frac{5}{2}\). \(\frac{2}{5}\times(-\frac{5}{2})=-1\), so they should be perpendicular? But wait Line 2 and Line 3 have the same slope, so they are parallel.

Wait, let's re - do:

1. Line 1 and Line 2:
\(m_1=\frac{2}{5}\), \(m_2 =-\frac{5}{2}\)
\(m_1\times m_2=\frac{2}{5}\times(-\frac{5}{2})=-1\) → Perpendicular.

2. Line 1 and Line 3:
\(m_1=\frac{2}{5}\), \(m_3=-\frac{5}{2}\)
\(m_1\times m_3=\frac{2}{5}\times(-\frac{5}{2})=-1\) → Perpendicular? Wait, no, wait Line 3's slope is \(-\frac{5}{2}\), same as Line 2. Wait, no, Line 1: \(\frac{2}{5}\), Line 3: \(-\frac{5}{2}\). The product is - 1, so they are perpendicular? But wait, Line 2 and Line 3 have the same slope, so they are parallel.

Wait, I think I made a mistake in Line 1's slope. Wait \(4x-10y=-6\)

\(-10y=-4x - 6\)

\(y=\frac{4x + 6}{10}=\frac{2x+3}{5}=\frac{2}{5}x+\frac{3}{5}\). Correct. So slope is \(\frac{2}{5}\).

Line 2: \(2y=-5x + 3\) → \(y=-\frac{5}{2}x+\frac{3}{2}\). Correct.

Line 3: \(y =-\frac{5}{2}x-8\). Correct.

So Line 1 and Line 2: perpendicular (product of slopes is - 1)

Line 1 and Line 3: \(m_1=\frac{2}{5}\), \(m_3 =-\frac{5}{2}\), product is - 1, so perpendicular? Wait, but Line 2 and Line 3 have the same slope, so they are parallel.

Wait, maybe I messed up the sign. Wait \(\frac{2}{5}\) and \(-\frac{5}{2}\): the product is - 1, so they are perpendicular. So Line 1 and Line 3: perpendicular? But Line 2 and Line 3: same slope, so parallel.

Line 1 and Line 2: perpendicular.

Line 1 and Line 3: perpendicular? Wait, no, wait Line 3's slope is \(-\frac{5}{2}\), same as Line 2. So Line 1 and Line 2: perpendicular, Line 1 and Line 3: perpendicular? But Line 2 and Line 3: parallel.

Wait, let's check with another approach. Two lines are parallel if their slopes are equal, perpendicular if the product of their slopes is - 1.

Line 1: \(m_1=\frac{2}{5}\)

Line 2: \(m_2=-\frac{5}{2}\)

Line 3: \(m_3 =-\frac{5}{2}\)

- Line 1 and Line 2: \(m_1\times m_2=\frac{2}{5}\times(-\frac{5}{2})=-1\) → Perpendicular.

- Line 1 and Line 3: \(m_1\times m_3=\frac{2}{5}\times(-\frac{5}{2})=-1\) → Perpendicular? Wait, no, \(\frac{2}{5}\) and \(-\frac{5}{2}\) are negative reciprocals, so their product is - 1, so they are perpendicular.

- Line 2 and Line 3: \(m_2=m_3=-\frac{5}{2}\) → Parallel.

Answer:

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