Three vertices of a parallelogram have coordinates (-2,2),(1,6) and (4,3). Find all possible coordinates of the fourth vertex. Do I use the distance formula for the two points and use them to add to each other to get two parallel sides? 

Here, Three points are given,
$P(-2,\;2)$
$Q(1,\;6)$
$R(4,\;3)$
and take $S(x,\;y)$
Since in parallelogram
$\vec{PS}=\vec{QR}$,
Therefore,
$$(Sx−Px,Sy−Py)=(Rx−Qx,Ry−Qy)$$
$$⇒(x−(−2),y−2)=(4−1,3−6)$$
$$⇒x−(−2)=4-1\;\;\implies x=0$$ 
 $$⇒y−2=3−6\;\;\implies y=-1$$
Thus, coordinates of fourth vertex are $S(0,\;-1)$
I hope it' ll help.
