The problem in the image is in Indonesian and asks the following:

"A parallelogram has 3 vertices at (1, 2), (3, 8), and (4, 1). Calculate the possible x-coordinate of the 4th vertex."

To solve this problem, we can use the property of parallelograms that the diagonals bisect each other. This means that the midpoint of one diagonal will coincide with the midpoint of the other diagonal.

Step-by-step solution:

1. Let the coordinates of the fourth vertex be $(x_4, y_4)$.

2. The midpoints of the diagonals should be the same: $\text{Midpoint of diagonal connecting } (1, 2) \text{ and } (x_4, y_4) = \text{Midpoint of diagonal connecting } (3, 8) \text{ and } (4, 1)$

   So, the midpoint formula gives: $\left(\frac{1 + x_4}{2}, \frac{2 + y_4}{2}\right) = \left(\frac{3 + 4}{2}, \frac{8 + 1}{2}\right)$ Simplifying the right-hand side: $\left(\frac{1 + x_4}{2}, \frac{2 + y_4}{2}\right) = (3.5, 4.5)$

3. Now, equate the corresponding x and y coordinates: $\frac{1 + x_4}{2} = 3.5 \quad \text{and} \quad \frac{2 + y_4}{2} = 4.5$

4. Solve for $x_4$ and $y_4$: $1 + x_4 = 7 \quad \Rightarrow \quad x_4 = 6$ $2 + y_4 = 9 \quad \Rightarrow \quad y_4 = 7$

Thus, the coordinates of the 4th vertex are $(6, 7)$.

Would you like more details on any step or have additional questions?

Related questions:

1. How do we find the area of a parallelogram given four vertices?
2. Can you explain the properties of a parallelogram in more detail?
3. How do we find the centroid of a parallelogram?
4. What other methods can be used to verify the fourth vertex of a parallelogram?
5. How can we generalize the formula for finding the 4th vertex of a parallelogram with arbitrary coordinates?

Tip:

For any geometric shape with vertices, the midpoint and slope formulas are often key tools to use, especially in problems involving parallelograms or triangles.