Question

The three vertices of a rhombus PQRS are P(2, – 3), Q(6, 5) and R(– 2, 1). Find the coordinates of the fourth vertex S and coordinates of the point where both the diagonals PR and QS intersect.

Hint:

For any parallelogram (including rectangles, rhombuses, and squares), the sum of opposite vertices' coordinates is equal:
\( x_P + x_R = x_Q + x_S \implies 2 + (-2) = 6 + x \implies x = -6 \).
\( y_P + y_R = y_Q + y_S \implies -3 + 1 = 5 + y \implies y = -7 \).
This vector addition shortcut finds the fourth vertex in just one step!

Answer 1

Step 1: Use the vector shortcut for a parallelogram's fourth vertex, instead of solving two midpoint equations.
A rhombus is a special parallelogram, and in any parallelogram $PQRS$ the diagonals $PR$ and $QS$ bisect each other, so their midpoints coincide, giving directly:
\[ P+R = Q+S \implies S = P+R-Q \]

Step 2: Substitute the known vertices directly into this single formula.
With $P(2,-3)$, $Q(6,5)$, $R(-2,1)$:
\[ S = (2,-3)+(-2,1)-(6,5) \]
x-coordinate: $2+(-2)-6=-6$. y-coordinate: $-3+1-5=-7$.
\[ S = (-6,-7) \]

Step 3: Find the point where the diagonals meet.
Since the diagonals bisect each other, the intersection is the midpoint of either diagonal. Using $P(2,-3)$ and $R(-2,1)$:
\[ M = \left(\frac{2+(-2)}{2}, \frac{-3+1}{2}\right) = (0,-1) \]

Final Answer:
The fourth vertex is $S(-6,-7)$, and the diagonals intersect at $(0,-1)$.
\[ \boxed{S(-6,-7),\ \text{intersection }(0,-1)} \]