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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 Parallelperpendicular Or Neither class=

Sagot :

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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