Answer:
Lines A and C are parallel
Lines B and D are perpendicular
Step-by-step explanation:
Given
See attachment for lines A to D
Required
Determine the parallel and perpendicular lines
To start with, we calculate the slope of each line.
Line A
The points on line A are:
[tex](x_1,y_1) = (3,-4)[/tex]
[tex](x_2,y_2) = (-1,2)[/tex]
Calculate the slope (m)
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m = \frac{2 - (-4)}{-1 - 3}[/tex]
[tex]m = \frac{2 +4}{-4}[/tex]
[tex]m = \frac{6}{-4}[/tex]
[tex]m_1 = -\frac{3}{2}[/tex]
Line B
[tex](x_1,y_1) = (-1,8)[/tex]
[tex](x_2,y_2) = (2,6)[/tex]
Calculate the slope (m)
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m = \frac{6 - 8}{2 -(-1)}[/tex]
[tex]m = \frac{-2}{2 +1}[/tex]
[tex]m_2 = -\frac{2}{3}[/tex]
Line C
[tex]y - 9 = -\frac{3}{2}(x + 8)[/tex]
Open bracket
[tex]y - 9 = -\frac{3}{2}x -\frac{3}{2}* 8[/tex]
[tex]y - 9 = -\frac{3}{2}x -3* 4[/tex]
[tex]y - 9 = -\frac{3}{2}x -12[/tex]
Make y the subject
[tex]y = -\frac{3}{2}x -12+9[/tex]
[tex]y = -\frac{3}{2}x -3[/tex]
The slope intercept of an equation is: [tex]y = mx + b[/tex]
Where
[tex]m = slope[/tex]
By comparison:
[tex]m_3 = -\frac{3}{2}[/tex]
Line D
[tex]12x - 8y = 40[/tex]
Subtract 12x from both sides
[tex]12x-12x - 8y = -12x+40[/tex]
[tex]- 8y = -12x+40[/tex]
Divide through by -8
[tex]\frac{- 8y}{-8} = \frac{-12x}{-8}+\frac{40}{-8}[/tex]
[tex]y= \frac{12x}{8}-\frac{40}{8}[/tex]
[tex]y= \frac{3}{2}x-5[/tex]
The slope intercept of an equation is: [tex]y = mx + b[/tex]
Where
[tex]m = slope[/tex]
By comparison:
[tex]m_4 = \frac{3}{2}[/tex]
So, the slopes of the lines are:
[tex]m_1 = -\frac{3}{2}[/tex] [tex]m_2 = -\frac{2}{3}[/tex] [tex]m_3 = -\frac{3}{2}[/tex] [tex]m_4 = \frac{3}{2}[/tex]
Lines with the same slope are parallel.
So:
Lines A and C with slope of [tex]-\frac{3}{2}[/tex] are parallel
Lines with the following relationship are perpendicular:
[tex]m * M = -1[/tex]
Lines B and D are perpendicular because:
[tex]-\frac{2}{3} * \frac{3}{2} = -1[/tex]
[tex]-\frac{6}{6} = -1[/tex]
[tex]-1 = -1[/tex]
