consider the line whose questionis 3x+y-2=0. then the slope of any line that is parallel to this line is ___?. the slope of any line that is perpendicular to this line is __.

Question

a1234 · Answer

Parallel lines have the same slope, but different y-intercepts. 

 3x+y-2=0

3x + y = 2

y = -3x + 2

The slope is -3.

anonymous · Answer

the slope of the parallel or perpendicular? its asking for both.

anonymous · Answer

to get the slope of a perp line
simply take the slope of that line, -3
change the numerator and denominator -1/3
and then change the sign 1/3
so the answer is
1/3

anonymous · Answer

or mathematically speaking
(slope of a line)(slope of the perp line for that line) = -1

anonymous · Answer

ok so it says " the slopeof any line that is perpendicular to this line__?

a1234 · Answer

Perpendicular lines have their slope producing a product of -1. 

The one I showed above is for parallel.

1/3*-3/1

= -3/3 = -1

Yes, 1/3.

anonymous · Answer

so -3 for the parallel and -1 for perp?

anonymous · Answer

1/3

anonymous · Answer

why you just want to copy the answer instead of trying to read and understand what we are typing

anonymous · Answer

we are spending our valuable time to help you understand something and you don't even care?
this is not how this site works

anonymous · Answer

I am trying to understand..

anonymous · Answer

i do cARE THAT IS WHY I AM TRYING TO UNDERSTAND
'

whpalmer4 · Answer

Okay, here's my best effort at helping you understand.

Original line is $$3x+y-2=0$$If we solve that for $y$, we put it in slope-intercept form, where it is easy to see the slope $(m)$:
$$3x+y-2=0$$$$3x+y=2$$$$y=-3x+2$$Compare that with the slope-intercept form:$$y=mx+b$$$$m = -3$$so our slope is $-3$ and any parallel line to this line will have a slope of $-3$.

Perpendicular lines (with one notable exception) have the property that the product of their slopes $= -1$.  Another name for this relationship is that they are negative reciprocals.  If the slope of the first line is $m_1$, then the slope of the second line is $$m_2 = -\frac{1}{m_1}$$Any line perpendicular to our line with slope $m = -3$ will have slope$$m = -\frac{1}{-3} = \frac{1}{3}$$

I said there was a notable exception to this relationship.  A vertical line $(x=c)$ has an undefined slope, and a horizontal line $(y=c)$ has a slope of 0.  ($c$ is a constant).  To construct the perpendicular line to a vertical or horizontal line, you have to write a new equation by hand in the appropriate form.  You can't calculate the new slope by taking a negative reciprocal.