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.
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.
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.