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Unit 4 linear equations homework 9 parallel and perpendicular lines (day1) activity
Directions if segments AB and CD are parallel,perpendicular, or neither
Unit 4 linear equations homework 9 parallel and perpendicular lines (day1) activity Directions if segments AB and CD are parallel,perpendicular, or neither

1 Answer

  • The lines that are parallel continuously have equal distances from each

    other and perpendicular lines form 90° angle at their intersection.

    The correct responses are;

    1. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥
    2. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel. ║
    3. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither
    4. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥
    5. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular, ⊥
    6. Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither
    7. The equations are parallel, ║
    8. The equations are perpendicular, ⊥
    9. The equations are perpendicular, ⊥
    10. The equations are neither parallel or perpendicular

    Reasons:

    Segment [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular

    [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel where; Slope of [tex]\overline{AB}[/tex] = Slope of [tex]\overline{CD}[/tex]

    [tex]\displaystyle \overline{AB} \ and \ \overline{CD} \ are \ perpendicular\ where; \ Slope \ of \ \overline{AB} = -\frac{1}{Slope \ of \ \overline{CD}}[/tex]

    [tex]1. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{\frac{13 - 3}{-2 - 0}} = \frac{10}{-2} = -5[/tex]

    [tex]\geq \displaystyle Slope \ of \ \overline{CD} = \frac{3 - 0}{10 - (-5)} = \frac{1}{5}[/tex]

    [tex]\displaystyle \mathbf{Slope \ of \ \overline{AB}} = -\frac{1}{Slope \ of \ \overline{CD}}[/tex], [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular

    [tex]2. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{\frac{7- 1}{3 - (-6)}} = \frac{6}{9} =\frac{2}{3}[/tex]

    [tex]\geq \displaystyle Slope \ of \ \overline{CD} = \frac{-5 - (-1)}{-6 - 0} = \frac{-4}{-6}=\frac{2}{3}[/tex]

    Slope of [tex]\mathbf{\overline{AB}}[/tex] = Slope of [tex]\overline{CD}[/tex] , [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel

    [tex]3. \ \displaystyle Slope \ of \ \overline{AB} = \mathbf{ \frac{2- 4}{-6 - (-2)} = \frac{-2}{-4}} =\frac{1}{2}[/tex]

    [tex]\displaystyle Slope \ of \ \overline{CD} = \frac{11 - (-7)}{-1 - 5} = \frac{18}{-6}=-3[/tex]

    [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither parallel or perpendicular

    [tex]4. \ \displaystyle Slope \ of \ \overline{AB} = \frac{8- 3}{-3 - 2} = \frac{5}{-5} =-1[/tex]

    [tex]\displaystyle Slope \ of \ \overline{CD} = \frac{6 - 2}{-4 - (-8)} = \frac{4}{4}=1[/tex]

    [tex]\displaystyle Slope \ of \ \overline{AB} = \mathbf{-\frac{1}{Slope \ of \ \overline{CD}}}[/tex], [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are perpendicular

    [tex]5. \ \displaystyle Slope \ of \ \overline{AB} = \frac{-1 - 2}{-8 - (-4)} = \frac{-3}{-4} =\frac{3}{4}[/tex]

    [tex]\displaystyle Slope \ of \ \overline{CD} = \mathbf{\frac{-3 - 6}{0 - 12}} = \frac{-9}{-12}=\frac{3}{4}[/tex]

    Slope of [tex]\overline{AB}[/tex] = Slope of [tex]\overline{CD}[/tex] , [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are parallel

    [tex]6. \ \displaystyle Slope \ of \ \overline{AB} = \frac{5 - (-1)}{6 - 3} = \frac{6}{3} =2[/tex]

    [tex]\displaystyle Slope \ of \ \overline{CD} = \mathbf{\frac{-5 - 7}{2 - (-4)}} = \frac{-12}{6}=-2[/tex]

    [tex]\overline{AB}[/tex] and [tex]\overline{CD}[/tex] are neither parallel or perpendicular

    Directions: To determine if the equations are parallel or perpendicular

    Solution:

    The equations are parallel if when expressed in the form, y = m·x  + c, the value of m are equal

    The equations are perpendicular, when we have;

    [tex]\displaystyle The \ value \ of \ \mathbf{m} \ in \ one \ equation = -\frac{1}{The \ value \ of \ \mathbf{m} \ in \ the \ other\ equation }[/tex]

    7. First equation

    3·x + 2·y = 6

    Therefore; [tex]\displaystyle y = -\frac{3}{2} \cdot x + 3[/tex]

    Second equation; [tex]\displaystyle y = \mathbf{-\frac{3}{2} \cdot x + 5}[/tex]

    The value of m which is the coefficient of x are both equal to [tex]\displaystyle -\frac{3}{2}[/tex]

    Therefore, the equations are parallel

    8.  Equation 1; 3·y = 4·x + 15

    Therefore;

    [tex]\displaystyle y = \frac{4}{3} \cdot x + 5[/tex]

    Equation 2; 9·x + 12·y = 12

    [tex]\displaystyle y = -\frac{9}{12} \cdot x + 1 = 1 - \frac{3}{4} \cdot x[/tex]

    The slope of the equation 2 is the negative inverse of the slope of equation 1, therefore, the equations are perpendicular

    9. Equation 1: 8·x - 2·y = 4

    Therefore;

    y = 4·x - 2

    Equation 2: x + 4·y = -12

    Therefore;

    [tex]\displaystyle y = -\frac{1}{4} \cdot x - 3[/tex]

    The slope of the equation 2 is the negative inverse of the slope of equation 1, therefore, the equations are perpendicular.

    10. Equation 1: 3·x + 2·y = 10

    Therefore;

    [tex]\displaystyle y = -\frac{3}{2} \cdot x - 5[/tex]

    Equation 2: 2·x + 3·y = -3

    [tex]\displaystyle y = -\frac{2}{3} \cdot x -1[/tex]

    The slopes of the equation 1 and 2 are not equal, and are not  the negative inverse of each other, therefore, the equations are neither parallel nor perpendicular.

    Learn more about parallel and perpendicular lines here:

    https://brainly.com/question/7197064

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