given a pair of lines
a parallel pair of lines is given whe the slope of both lines is the same or
[tex]m1=m2[/tex]
perpendicular is given when slope1 multiply by slope2 equals -1 or
[tex]m1*m2=-1[/tex]
first pair.
[tex]2x-4y=-4[/tex]
solving for y
[tex]y=\frac{4+2x}{4}=\frac{x+2}{2}[/tex][tex]y=\frac{1}{2}x+1[/tex]
then m1=1/2
[tex]3x-6y=9[/tex]
solving for y
[tex]-6y=9-3x[/tex][tex]y=\frac{x-3}{2}[/tex][tex]y=\frac{1}{2}x-\frac{3}{2}[/tex]
then m2=1/2
since both slopes are equal, first pair has paralel lines
2nd pair
[tex]2x+3y=12[/tex][tex]y=\frac{12-2x}{3}[/tex][tex]y=-\frac{2}{3}x+4[/tex]
then m1= -2/3
[tex]5x-4y=7[/tex][tex]y=\frac{-7+5x}{4}[/tex][tex]y=\frac{5}{4}x-\frac{7}{4}[/tex]
m2 = 5/4
since
[tex]m1*m2=-1[/tex][tex]-\frac{2}{3}*\frac{5}{4}=\frac{10}{12}[/tex]
second pair is neither parallel nor perpendicular
3rd pair
[tex]x+\frac{1}{2}y=2[/tex][tex]y=2(2-x)[/tex][tex]y=-2x+4[/tex]
then m1 = -2
[tex]\frac{1}{2}x+2y=4[/tex][tex]2y=4-\frac{1}{2}x[/tex][tex]y=-\frac{1}{4}x+2[/tex]
then m2=-1/4
[tex]m1*m2=-2*-\frac{1}{4}=2[/tex]
3rd pair is not a paralel line nor perpendicular line
