Geometry help, please!!! Determine if the lines are parallel, perpendicular, or neither. #1) y=1/2x+7 y-8=-2(x+9)
@Hero
1. Put the second equation in the form \(y = mx + b\) 2. Test the slopes using \(m_1m_2 = -1\) If their products = -1 then the lines are perpendicular.
y=1/2x+7 and y-8=-2(x+9) where do i go from here?
I just told you above what to do. Follow the steps.
I'm unsure of what to plug in
Like Hero said, isolate y \(y-8=-2(x+9)\) First we want to distribute to make this easier, can you do that?
y-8=-2(x+9) y-8=-2x+18 y=-2x+10 is this right?
How come you didn't do that when I asked?
i didn't know what you meant when you said m-m1
$m_1$ and $m_2$ are the slopes from each line.
Not quite, you have to do \(-2 \times x \) + \(-2\times 9\) \(-2 \times x \) is -2x and \(-2\times 9\) is -18 (You missed the sign transfer) Then you have to add 8 on both sides, right idea \(y-8=-2x-18\) ==> \(y=-2x-10\) Keep the sign and you're good
All that matters really are the slopes.
You have them now multiply them together. If you get -1 then the lines are perpendicular.
Well true but I just want to show the whole process, shortening it might be confusing
I was unsure of the steps, thanks though Hero.
y=-2x-10 times y=1/2x+7 is y=-2x+-70 ????
Hero said to multiply the slopes
y=-1x+-70 ***
oh
so it is perpendicular, it =-1
Yep
https://www.desmos.com/calculator/mg3bxn27zh Graphically (Not sure if you were allowed to use graphing calculators)
yeah, can't use graphing calcs
i have another, can i work it out and you check it?
Sure
y+3=5(x-4) y-2=5(x+2) y-2=5(x+2) y+3=5x-20 y=5x+23 -- y-2=5(x+2) y-2=5x+10 y=5x-8 -- 5*5=25 ?
I'm confused on your work, what did you do
If I'm reading this right \(y+3=5x-20\\ y=5x+23\) Watch your sign -20-3 ≠ [+]23
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