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SephI
Help with graphing, please :P
Choose the slope-intercept equation of the line that passes through the point (-5, -1) and is perpendicular to y = 5/2x + 2
They have the same slope since they are perpendicular. so I know (-5, -1) with a slope of 5/2 has something to do with the equation, but what happens next?
they dont have the same slope, that is when they are parallel. For perpendicular the slope is opposite
its a negative reciprocal, so can you figure it out what the slope is from there?
@SephI User Please Respond.
Oh, ok. So I would have a slope of 2/5, and then what?
close.. m=-2/5 then you use your point-interecpt form: y-y=m(x-x)
y - (-1) = -2/5[(x - (-5)] y + 1 = -2/5x + 5 y = -2/5x - 4? ^ That's not one of my options. So I did it wrong...
thats close but you need to distribute the (-2/5) to the 5 as well
that is because the x+5 is in parentheses. so the -2/5 applies to both pieces
or (x-(-5)) however you want to think about it
y - (-1) = -2/5[(x - (-5)] y - (-1) = -2/5x + ??? How would I distribute the 5? Wouldn't it be -2/5*5 Which is -10/5 = -2?
yes! perfect. so you would have (y-(-1))=-2/5x-2 then you just need to y by itself
y + 1 = -2/5x - 2 y = -2/5x - 3 - 1 y = -2/5x - 3?
yep! That should be your answer.