Answer :
Sure! Let's determine whether the lines are parallel, perpendicular, or neither for each exercise by calculating the slopes and comparing them.
### Exercise 63
For [tex]\(L_1\)[/tex] passing through points [tex]\((0, -1)\)[/tex] and [tex]\((5, 9)\)[/tex]:
- The slope of a line passing through points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by [tex]\(m = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex].
The slope of [tex]\((0, -1)\)[/tex] and [tex]\((5, 9)\)[/tex]:
[tex]\[ m_{L1} = \frac{9 - (-1)}{5 - 0} = \frac{10}{5} = 2.0 \][/tex]
Since there is only one line present, there is no comparison to be made.
Result for 63:
Slope of [tex]\(L_1\)[/tex]: 2.0
Relationship: Only one line present
### Exercise 64
For [tex]\(L_1\)[/tex] passing through points [tex]\((−2, −1)\)[/tex] and [tex]\((1, 5)\)[/tex]:
[tex]\[ m_{L1} = \frac{5 - (-1)}{1 - (-2)} = \frac{6}{3} = 2.0 \][/tex]
For [tex]\(L_2\)[/tex] passing through points [tex]\((0, 3)\)[/tex] and [tex]\((4, 1)\)[/tex]:
[tex]\[ m_{L2} = \frac{1 - 3}{4 - 0} = \frac{-2}{4} = -0.5 \][/tex]
We compare the slopes:
[tex]\[ m_{L1} = 2.0 \][/tex]
[tex]\[ m_{L2} = -0.5 \][/tex]
Since [tex]\(m_{L1} \times m_{L2} = 2.0 \times -0.5 = -1\)[/tex], [tex]\(L_1\)[/tex] and [tex]\(L_2\)[/tex] are perpendicular.
Result for 64:
Slope of [tex]\(L_1\)[/tex]: 2.0
Slope of [tex]\(L_2\)[/tex]: -0.5
Relationship: Perpendicular
### Exercise 65
For [tex]\(L_1\)[/tex] passing through points [tex]\((3, 6)\)[/tex] and [tex]\((-6, 0)\)[/tex]:
[tex]\[ m_{L1} = \frac{0 - 6}{-6 - 3} = \frac{-6}{-9} = \frac{2}{3} \][/tex]
Since there is only one line present, there is no comparison to be made.
Result for 65:
Slope of [tex]\(L_1\)[/tex]: 0.6666666666666666 (or [tex]\(\frac{2}{3}\)[/tex])
Relationship: Only one line present
### Exercise 66
For [tex]\(L_1\)[/tex] passing through points [tex]\((4, 8)\)[/tex] and [tex]\((-4, 2)\)[/tex]:
[tex]\[ m_{L1} = \frac{2 - 8}{-4 - 4} = \frac{-6}{-8} = 0.75 \][/tex]
For [tex]\(L_2\)[/tex] passing through points [tex]\((0, -1)\)[/tex] and [tex]\(\left(5, \frac{7}{3}\right)\)[/tex]:
[tex]\[ m_{L2} = \frac{\frac{7}{3} - (-1)}{5 - 0} = \frac{\frac{7}{3} + \frac{3}{3}}{5} = \frac{\frac{10}{3}}{5} = \frac{10}{3 \times 5} = \frac{10}{15} = \frac{2}{3} = 0.6666666666666667 \][/tex]
Comparing slopes:
[tex]\[ m_{L1} = 0.75 \][/tex]
[tex]\[ m_{L2} = 0.6666666666666667 \][/tex]
Since [tex]\(m_{L1} \neq m_{L2}\)[/tex] and [tex]\(m_{L1} \times m_{L2} \neq -1\)[/tex], [tex]\(L_1\)[/tex] and [tex]\(L_2\)[/tex] are neither parallel nor perpendicular.
For the second [tex]\(L_2\)[/tex] passing through points [tex]\((3, -5)\)[/tex] and [tex]\((−1, \frac{1}{3})\)[/tex]:
[tex]\[ m_{L2\_second} = \frac{\frac{1}{3} - (-5)}{-1 - 3} = \frac{\frac{1}{3} + 5}{-4} = \frac{\frac{1 + 15}{3}}{-4} = \frac{\frac{16}{3}}{-4} = -\frac{4}{3} = -1.3333333333333333 \][/tex]
We compare the slopes again:
[tex]\[ m_{L1} = 0.75 \][/tex]
[tex]\[ m_{L2\_second} = -1.3333333333333333 \][/tex]
Since [tex]\(m_{L1} \times m_{L2\_second} = 0.75 \times -1.3333333333333333 = -1\)[/tex], [tex]\(L_1\)[/tex] and the second [tex]\(L_2\)[/tex] are perpendicular.
Result for 66:
Slope of [tex]\(L_1\)[/tex]: 0.75
Slope of [tex]\(L_2\)[/tex]: 0.6666666666666667
Relationship: Neither
Slope of the second [tex]\(L_2\)[/tex]: -1.3333333333333333
Relationship with second [tex]\(L_2\)[/tex]: Perpendicular
### Exercise 63
For [tex]\(L_1\)[/tex] passing through points [tex]\((0, -1)\)[/tex] and [tex]\((5, 9)\)[/tex]:
- The slope of a line passing through points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by [tex]\(m = \frac{y_2 - y_1}{x_2 - x_1}\)[/tex].
The slope of [tex]\((0, -1)\)[/tex] and [tex]\((5, 9)\)[/tex]:
[tex]\[ m_{L1} = \frac{9 - (-1)}{5 - 0} = \frac{10}{5} = 2.0 \][/tex]
Since there is only one line present, there is no comparison to be made.
Result for 63:
Slope of [tex]\(L_1\)[/tex]: 2.0
Relationship: Only one line present
### Exercise 64
For [tex]\(L_1\)[/tex] passing through points [tex]\((−2, −1)\)[/tex] and [tex]\((1, 5)\)[/tex]:
[tex]\[ m_{L1} = \frac{5 - (-1)}{1 - (-2)} = \frac{6}{3} = 2.0 \][/tex]
For [tex]\(L_2\)[/tex] passing through points [tex]\((0, 3)\)[/tex] and [tex]\((4, 1)\)[/tex]:
[tex]\[ m_{L2} = \frac{1 - 3}{4 - 0} = \frac{-2}{4} = -0.5 \][/tex]
We compare the slopes:
[tex]\[ m_{L1} = 2.0 \][/tex]
[tex]\[ m_{L2} = -0.5 \][/tex]
Since [tex]\(m_{L1} \times m_{L2} = 2.0 \times -0.5 = -1\)[/tex], [tex]\(L_1\)[/tex] and [tex]\(L_2\)[/tex] are perpendicular.
Result for 64:
Slope of [tex]\(L_1\)[/tex]: 2.0
Slope of [tex]\(L_2\)[/tex]: -0.5
Relationship: Perpendicular
### Exercise 65
For [tex]\(L_1\)[/tex] passing through points [tex]\((3, 6)\)[/tex] and [tex]\((-6, 0)\)[/tex]:
[tex]\[ m_{L1} = \frac{0 - 6}{-6 - 3} = \frac{-6}{-9} = \frac{2}{3} \][/tex]
Since there is only one line present, there is no comparison to be made.
Result for 65:
Slope of [tex]\(L_1\)[/tex]: 0.6666666666666666 (or [tex]\(\frac{2}{3}\)[/tex])
Relationship: Only one line present
### Exercise 66
For [tex]\(L_1\)[/tex] passing through points [tex]\((4, 8)\)[/tex] and [tex]\((-4, 2)\)[/tex]:
[tex]\[ m_{L1} = \frac{2 - 8}{-4 - 4} = \frac{-6}{-8} = 0.75 \][/tex]
For [tex]\(L_2\)[/tex] passing through points [tex]\((0, -1)\)[/tex] and [tex]\(\left(5, \frac{7}{3}\right)\)[/tex]:
[tex]\[ m_{L2} = \frac{\frac{7}{3} - (-1)}{5 - 0} = \frac{\frac{7}{3} + \frac{3}{3}}{5} = \frac{\frac{10}{3}}{5} = \frac{10}{3 \times 5} = \frac{10}{15} = \frac{2}{3} = 0.6666666666666667 \][/tex]
Comparing slopes:
[tex]\[ m_{L1} = 0.75 \][/tex]
[tex]\[ m_{L2} = 0.6666666666666667 \][/tex]
Since [tex]\(m_{L1} \neq m_{L2}\)[/tex] and [tex]\(m_{L1} \times m_{L2} \neq -1\)[/tex], [tex]\(L_1\)[/tex] and [tex]\(L_2\)[/tex] are neither parallel nor perpendicular.
For the second [tex]\(L_2\)[/tex] passing through points [tex]\((3, -5)\)[/tex] and [tex]\((−1, \frac{1}{3})\)[/tex]:
[tex]\[ m_{L2\_second} = \frac{\frac{1}{3} - (-5)}{-1 - 3} = \frac{\frac{1}{3} + 5}{-4} = \frac{\frac{1 + 15}{3}}{-4} = \frac{\frac{16}{3}}{-4} = -\frac{4}{3} = -1.3333333333333333 \][/tex]
We compare the slopes again:
[tex]\[ m_{L1} = 0.75 \][/tex]
[tex]\[ m_{L2\_second} = -1.3333333333333333 \][/tex]
Since [tex]\(m_{L1} \times m_{L2\_second} = 0.75 \times -1.3333333333333333 = -1\)[/tex], [tex]\(L_1\)[/tex] and the second [tex]\(L_2\)[/tex] are perpendicular.
Result for 66:
Slope of [tex]\(L_1\)[/tex]: 0.75
Slope of [tex]\(L_2\)[/tex]: 0.6666666666666667
Relationship: Neither
Slope of the second [tex]\(L_2\)[/tex]: -1.3333333333333333
Relationship with second [tex]\(L_2\)[/tex]: Perpendicular
