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

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;

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.

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