If P (4, 10), 9 (9, 5), R(1, 4) and S (x, 9) are the vertices of parallelogram PQRS. then x...........

1. Break Down the Problem

We are given three vertices of a parallelogram: P(4, 10), Q(9, 5), R(1, 4), and we need to find the x-coordinate of vertex S(x, 9). In a parallelogram, the midpoints of the diagonals bisect each other. Therefore, we can use the properties of midpoints to find S.

2. Relevant Concepts

The midpoint $M$ of two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ For diagonals PR and QS, their midpoints should be equal.

3. Analysis and Detail

Let's find the midpoint of diagonal PR:

• Coordinates of P: $(4, 10)$
• Coordinates of R: $(1, 4)$

Calculating the midpoint $M_{PR}$: $M_{PR} = \left( \frac{4 + 1}{2}, \frac{10 + 4}{2} \right) = \left( \frac{5}{2}, 7 \right)$

Now, we know that the midpoint of diagonal QS must equal $M_{PR}$. Let’s calculate the midpoint of Q(9, 5) and S(x, 9): $M_{QS} = \left( \frac{9 + x}{2}, \frac{5 + 9}{2} \right) = \left( \frac{9 + x}{2}, 7 \right)$

4. Verify and Summarize

Since $M_{PR} = M_{QS}$, we can equate their x-coordinates: $\frac{9 + x}{2} = \frac{5}{2}$ To solve for $x$:

1. Multiply both sides by 2: $9 + x = 5$
2. Subtract 9 from both sides: $x = 5 - 9 = -4$

Final Answer

Therefore, the x-coordinate of vertex S is $x = -4$.