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They cannot be parallel and so, since the question has been asked, as is clearly part of your homework, it must have an answer other than "random lines". So then, if you assume that the lines are
y = 2x + 4
and
x + 2y = 12
they are mutually perpendicular.
If: 12 = 2y+x then y = -1/2x+6 So: y = 2x+4 and y = -1/2x+6 which means that they are perpendicular lines
They are parallel lines.
x + 2y = 7 2y = -x + 7 y = -(x-7)/2 => -x/2 + 7/2 2x - y = 7 -y = -2x + 7 y = 2x + 7 Since -1/2 is the negative reciporical of 2, the slopes of these equations are perpendicular. Therefore, these two lines are perpendicular.
Perpendicular
6x-2x-6xy-2x-2y4x-2y=2x-2y-14xy
If: 12 = 2y+x then y = -1/2x+6 So: y = 2x+4 and y = -1/2x+6 which means that they are perpendicular lines
They are straight lines.
It is not one, but they are two perpendicular lines.
perpendicular JEW
They are parallel lines.
-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x
If you mean 3x+2y = -5 and -2x+3y = -5 then they are straight line equations
If you mean: y = 2x and 2y = -x+4 then they are positive and negative straight line equations
parallel
2x+y=-12-4x-2y=30 x=-y/2+30, you plug this into -12-4x-2y=30 and get -12+2y-120-2y=30 which gives 2y-2y=162. This is not possible, so the equation is unsolvable.
Without any equality signs the given terms can't be considered to be equations.
x + 2y = 7 2y = -x + 7 y = -(x-7)/2 => -x/2 + 7/2 2x - y = 7 -y = -2x + 7 y = 2x + 7 Since -1/2 is the negative reciporical of 2, the slopes of these equations are perpendicular. Therefore, these two lines are perpendicular.