1. ## Vectors problem.

ABCD is a parallelogram, where A is (4,2), B is (-6,1) and D is (-3,-4). Find the coordinates of C.

My attempt:
$\displaystyle \vec{AC} = \vec{BD}$
First I'll find the vector $\displaystyle \vec{BD}$ then solve for its magnitude:
$\displaystyle \bec{BD} = \vec{OD} - \vec{OB} => (-3-(-6),(-4-1) => (3,-5)$
Now I'll find the magnitude:
$\displaystyle \vec{BD} = \sqrt{(3)^{2}+(-5)^{2}}$
$\displaystyle \vec {BD} = \sqrt{34}$
Unfortunately, this is the farthest I can get. I tried isolating for $\displaystyle \vec{C}$ but didn't work out as I only got 1 value and not two. I don't have the slightest clue on how to solve this problem (never saw it in my work or problems) and it threw me off completely. Thanks in advance.

2. Originally Posted by Pupil
ABCD is a parallelogram, where A is (4,2), B is (-6,1) and D is (-3,-4). Find the coordinates of C.
let the coordinates of $\displaystyle C$ be $\displaystyle (x,y)$

$\displaystyle \vec{AB} = \vec{DC}$

$\displaystyle <-10,-1> = <x-(-3), y-(-4)>$

$\displaystyle C = (-13,-5)$