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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?

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)

-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?

oh I see what went wrong

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)

I'm glad I was of help