Find the 4th vertex of the parallelogram

You are given that $P(-2;4)$, $Q(1;-2)$ and $R(3;3)$ and $S(x;y)$ are vertices of parallelogram $PQRS$. Given that $PQRS$ is a parallelogram, determine the coordinates of $S$.

Please can you show me how to find $S$ from the information given.

• What properties of a parallelogram do you know that could be helpfull in a situation like this? – gebruiker Aug 4 '14 at 12:56
• Perhaps you are specifying the order of the vertices as they appear in the parallelogram as $PQRS$, which would remove ambiguity. In general having three vertices of a parallelogram allows for three possible positions of the fourth vertex, depending on which pair of the three are endpoints of a diagonal rather than a side. – hardmath Aug 4 '14 at 13:00
• Just a small quibble: "vertice" is not an English word nor a term for writing mathematics in English. A single element from a set of vertices is a vertex. – David K Aug 4 '14 at 13:03

Since $\vec{PS}=\vec{QR}$, you'll have $$(S_x-P_x,S_y-P_y)=(R_x-Q_x,R_y-Q_y)$$ $$\Rightarrow (x-(-2),y-4)=(3-1,3-(-2))$$ $$\Rightarrow x-(-2)=3-1\ \text{and}\ y-4=3-(-2)\Rightarrow S(0,9).$$

• Slightly faster if you just stick to vectors! It's equivalent though. – rrogers Sep 16 at 0:49

The two diagonals of a parallelogram bisect each other. So the midpoint of $PR$ is the midpoint of $QS$.

Compute the midpoint of $PR$, call it $M$.

Compute the midpoint of $QS$. Since the coordinates of $S$ are as yet unknown, the result of this calculation is a point whose coordinates are arithmetic expressions involving the variables $x$ and $y$.

Now you have two ways of writing the coordinates of the same point, which gives you two equations (first coordinate equals first coordinate, etc.).

That should be sufficient to find the answer.

Hint:

Since we're talking of a paralelogram, $Q-P = R-S$ and $P-S = Q-R$, i.e:

$$(1,-2) - (-2,4) = (3,-6) = (3,3) - (x,y)\\ (1,-2) - (3,3) = (-2,-5) = (-2,4) - (x,y)$$