# Vectors problem.

• Mar 2nd 2011, 03:22 PM
Pupil
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.
• Mar 2nd 2011, 03:58 PM
skeeter
Quote:

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 $\displaystyle (x,y)$

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

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

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