Coordinates for D. (?,6) C is (10,0) Rectangle ABCD. A(2,14) B(-2,8) C(10,0) D(?,6)

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Coordinates for D. (?,6) C is (10,0) Rectangle ABCD. A(2,14) B(-2,8) C(10,0) D(?,6)

Mathematics
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earlier we found line BC to be \[\Large y = -\frac{2}{3}x+\frac{20}{3}\]
Yes, would it be the same for AD?

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Other answers:

no
but the same steps will be used find the slope of AD (since it's parallel to BC, the two slopes are equal). Then use it to find the equation of line AD
you also need the equation for line CD
For AD equation, would it start like: y-14=m(x-2)? m = slope
yes and m = -2/3 because AD and BC have the same slope (they are parallel lines)
So, then I can plug in 6 to get the x value for D?
not quite
you'll need equation CD
oh wait, nvm
yes you can plug in y = 6. I didn't see D(?,6)
Yeah, sorry, with my packet, I have image of it showing the C point being on 0. So, helped me determine the distance for the AD y and what to subtract.
-2/3 * -2 would be 12/3?
-2/3 * -2 = 4/3
6-14=-2/3(x-2) -8=-2/3x + 4/3 It ends up being: 6 = -2/3x + 28/3 right?
good
now solve for x
x=5?
no
oh I see what went wrong
When you went from -8=-2/3x + 4/3 to 6 = -2/3x + 28/3 you made an error the 6 in ` 6 = -2/3x + 28/3` should be 0 so you should have 0 = -2/3x + 28/3
it's 0 because you added 8 to both sides the -8 and +8 combine to 0 on the left side
OH. x=14
yep D is (14,6)
To be honest, I read my graph wrong aswell, so I was confused. I lack in graphing, but all is well. You really helped and taught me a few things, I hope to remember. (:
I'm glad I was of help

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