Determine if the lines are parallel, perpendicular or neither for -y=3x-2 and -6x+2y=6

To determine if two lines are parallel or perpendicular, you have to look at the slopes. For the line y=kx+b, k is the slope.

For two lines  y=k₁x+b₁ and y=k₂X+b_{2 ,}When K₁=K₂, two lines are parallel; when K₁=-1/k₂, two lines are perpendicular. It doesn't matter whatever b₁, b₂ are. Hence to save time, you only need to calculate  k₁ and k_2.

To calculate the slope k of any line, you have to change the equation to y=kx+b

For -y=3x-2, k=3/(-1)=-3 (you divide -1 on each side of =, but you don't need to calculate -2/(-1))

For -6X+2y=6, K=6/2=3 (you first move the -6x to right side, it becomes 6x, then divide by 2)

Now you can get the answer: The two lines neither parallel nor perpendicular.

First, we need to get the equations in the form y = mx + b. For -y=3x-2, multiply both sides by -1, to give:

-(-y) = -(3x - 2)   ====>   y = = -3x + 2

for -6x+2y=6, we need to add 6x to both sides:
-6x + 6x + 2y = 6 + 6x    =====>   2y = 6 + 6x
and then divide both sides by 2
2y/2 = (6 + 6x)/2       =====>        y = 3x + 3

Now, we have y = -3x + 2 and y = 3x + 3. If the lines are parallel, they will have the same slope/gradient, which is the m in the form y = mx + b. If they are perpendicular, the the gradients will be related by the equation m₁ = -1/m₂ or m₂ = -1/m₁. 

They aren't parallel, as -3 ≠ 3, nor they aren't perpendicular, as -3 ≠ -1/3, or 3 ≠ -1/-3.

As a first step, convert both equations to slope-intecept form by solving for y.

First equation:

-y = 3x - 2  Multiply by -1

y = -3x + 2

Second equation:

-6x + 2y = 6 Add 6x to each side

2y = 6x + 6  Divide both sides by 2

y = 3x + 2

To determine if two lines are parallel, first look at the slopes (coefficient on x).  If the slopes are the same, then the lines are either parallel or the same line (dependent).  In the case of these lines, the slopes are not the same, so they are not parallel.

To determine if two lines are perpendicular, look at the slopes.  If the slopes are inverse (fractions where the numerators and denominators are switched) opposites (one is positive and one is negative), then the lines are perpendicular.  In more basic terms, if the slopes were 3 and -1/3, they would be perpendicular.  Since they are not inverses, they are not perpendicular.

The lines are neither parallel nor perpendicular.