Question

suppose we are given the following.
line 1 passes through (-7, -2) and (0, 3).
line 2 passes through (-5, 8) and (0, 1).
line 3 passes through (7, 2) and (0, -3).
(a) find the slope of each line.
slope of line 1:
slope of line 2:
slope of line 3:
(b) 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: Recall slope formula

The slope \( m \) of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by \( m=\frac{y_2 - y_1}{x_2 - x_1} \).

Step2: Calculate slope of Line 1

Line 1 passes through \((-7, -2)\) and \((0, 3)\). Let \( (x_1, y_1)=(-7, -2) \) and \( (x_2, y_2)=(0, 3) \). Then slope \( m_1=\frac{3 - (-2)}{0 - (-7)}=\frac{3 + 2}{0 + 7}=\frac{5}{7} \).

Step3: Calculate slope of Line 2

Line 2 passes through \((-5, 8)\) and \((0, 1)\). Let \( (x_1, y_1)=(-5, 8) \) and \( (x_2, y_2)=(0, 1) \). Then slope \( m_2=\frac{1 - 8}{0 - (-5)}=\frac{-7}{5} \).

Step4: Calculate slope of Line 3

Line 3 passes through \((7, 2)\) and \((0, -3)\). Let \( (x_1, y_1)=(7, 2) \) and \( (x_2, y_2)=(0, -3) \). Then slope \( m_3=\frac{-3 - 2}{0 - 7}=\frac{-5}{-7}=\frac{5}{7} \).

Step5: Analyze Line 1 and Line 2

Two lines are parallel if their slopes are equal, and perpendicular if the product of their slopes is \(-1\). For Line 1 (\( m_1=\frac{5}{7} \)) and Line 2 (\( m_2=-\frac{7}{5} \)), the product \( m_1\times m_2=\frac{5}{7}\times(-\frac{7}{5})=-1 \). So they are perpendicular.

Step6: Analyze Line 1 and Line 3

For Line 1 (\( m_1=\frac{5}{7} \)) and Line 3 (\( m_3=\frac{5}{7} \)), since \( m_1 = m_3 \), they are parallel.

Step7: Analyze Line 2 and Line 3

For Line 2 (\( m_2=-\frac{7}{5} \)) and Line 3 (\( m_3=\frac{5}{7} \)), the product \( m_2\times m_3=-\frac{7}{5}\times\frac{5}{7}=-1 \)? Wait no, \( -\frac{7}{5}\times\frac{5}{7}=-1 \)? Wait \( -\frac{7}{5}\times\frac{5}{7}=-1 \)? Wait no, \( \frac{5}{7}\times(-\frac{7}{5})=-1 \), but here \( m_2=-\frac{7}{5} \) and \( m_3=\frac{5}{7} \), product is \( -1 \)? Wait no, wait \( m_2=-\frac{7}{5} \), \( m_3=\frac{5}{7} \), product is \( -\frac{7}{5}\times\frac{5}{7}=-1 \)? Wait no, \( \frac{5}{7}\times(-\frac{7}{5})=-1 \), yes. Wait but Line 1 and Line 3 have same slope, Line 2 and Line 3: \( m_2=-\frac{7}{5} \), \( m_3=\frac{5}{7} \), product is \( -1 \), so they are perpendicular? Wait no, wait let's recalculate. Wait Line 2 slope: \( (1 - 8)/(0 - (-5))=( -7)/5=-\frac{7}{5} \). Line 3 slope: \( (-3 - 2)/(0 - 7)=(-5)/(-7)=\frac{5}{7} \). Then \( m_2\times m_3=(-\frac{7}{5})\times(\frac{5}{7})=-1 \). Wait but Line 1 and Line 3 are parallel (same slope), Line 1 and Line 2 are perpendicular (product -1), Line 2 and Line 3: product of slopes is \( -\frac{7}{5}\times\frac{5}{7}=-1 \), so they are perpendicular? Wait no, wait maybe I made a mistake. Wait Line 3 slope: ( -3 - 2 ) / (0 - 7 ) = (-5)/(-7)=5/7. Correct. Line 2 slope: (1 - 8)/(0 - (-5))=(-7)/5=-7/5. So \( m_2\times m_3=(-7/5)\times(5/7)=-1 \), so they are perpendicular. Wait but let's check again.

Wait for part (a):

Slope of Line 1: \( \frac{3 - (-2)}{0 - (-7)}=\frac{5}{7} \)

Slope of Line 2: \( \frac{1 - 8}{0 - (-5)}=\frac{-7}{5} \)

Slope of Line 3: \( \frac{-3 - 2}{0 - 7}=\frac{-5}{-7}=\frac{5}{7} \)

Part (b):

Line 1 and Line 2: slopes \( \frac{5}{7} \) and \( -\frac{7}{5} \), product is \( -1 \), so perpendicular.

Line 1 and Line 3: slopes \( \frac{5}{7} \) and \( \frac{5}{7} \), so parallel.

Line 2 and Line 3: slopes \( -\frac{7}{5} \) and \( \frac{5}{7} \), product is \( -1 \), so perpendicular? Wait no, wait \( -\frac{7}{5}\times\frac{5}{7}=-1 \), yes. So they are perpendicular.

Answer:

(a) Slope of Line 1: \(\frac{5}{7}\), Slope of Line 2: \(-\frac{7}{5}\), Slope of Line 3: \(\frac{5}{7}\)

(b) Line 1 and Line 2: Perpendicular; Line 1 and Line 3: Parallel; Line 2 and Line 3: Perpendicular