# Determine if the line y+x-2 = 0 and the line 6y-3x +8 = 0 are parallel, perpendicular, or neither.

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Given the lines:

y + x -2 = 0

6y-3x + 8 =0

We need to determine the relation between the lines ( parallel, perpendicular, or neither)

First we will use the slope to find the relation.

If the slopes are equal, then the lines are parallel.

if the product of the slopes is -1, then the slope are perpendicular.

Let us rewrite the equations of the lines into the slope form.

==> y = -x + 2 ............(1)

==> y = (3/6)x - 8/6

==> y= (1/2)x - 4/3........(2)

We notice that the slopes are not equal. then the lines are not parallel.

Also, the product of the slopes is -1*1/2 = -1/2 , then the lines are not perpendicular.

**Then the slopes are not parallel nor perpendicular.**

We have the lines y+x-2 = 0 and 6y-3x +8 = 0

y+x-2 = 0

=> y = -x + 2

This is of the form y = mx + c where m is the slope.

So the slope is -1

6y-3x +8 = 0

=> 6y = 3x - 8

=> y = x/2 - 8/6

The slope is 1/2

Now -1 and 1/2 are neither equal nor negative reciprocals.

**Therefore the two lines are neither parallel nor perpendicular.**

The lines a1x+b1y+c1= 0 and a2x+b2y+c2 = 0 are parallel if b1/a1 = b2/a2 , perpendicular if b1/a1 = - a1/b2, and the lines are not parallel and not perpendicular other wise.

given lines are y+x-2 = 0 and 6y-3x +8 = 0.

Or

x+y-2 = 0 and 3x-6y -8 = 0.

b1/a1 = 1/1 = 1.

b2/a2 = -6/3 = -2.

So b1/a1 not equal to b2/a2 . So the lines are not parallel.

b1/a1 = 1 and -a2/b2 = -(-3/6) = 1/2. So b1/a1 = not equal to -a2/b2 . So the lines are not perpendicular.

Therefore the lines are neither parallel nor perpendicular.