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?
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$\begingroup$Since $\vec{PS}=\vec{QR}$, you'll have $$(S_x-P_x,S_y-P_y)=(R_x-Q_x,R_y-Q_y)$$ $$\Rightarrow (x-(-2),y-4)=(3-1,3-(-2))$$ $$\Rightarrow x-(-2)=3-1\ \text{and}\ y-4=3-(-2)\Rightarrow S(0,9).$$
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$Hint:
Since we're talking of a paralelogram, $Q-P = R-S$ and $P-S = Q-R$, i.e:
$$(1,-2) - (-2,4) = (3,-6) = (3,3) - (x,y)\\ (1,-2) - (3,3) = (-2,-5) = (-2,4) - (x,y)$$
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