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.**

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.**

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