The problem involves finding the coordinates of a weighted average point given two points and their respective weights. From the image, I can see:

• The two points on the graph:

  1. $P_1 = (-2, 1)$
  2. $P_2 = (2, 5)$

• The weights for these points:

  • Point $P_1$ has a weight of 4.
  • Point $P_2$ has a weight of 1.

Formula for the Weighted Average:

The coordinates $(x, y)$ of the weighted average point are calculated using the following formula: $x_{\text{avg}} = \frac{w_1 \cdot x_1 + w_2 \cdot x_2}{w_1 + w_2}$ $y_{\text{avg}} = \frac{w_1 \cdot y_1 + w_2 \cdot y_2}{w_1 + w_2}$ Where:

• $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the points.
• $w_1$ and $w_2$ are the respective weights of the points.

Substituting the values:

• For the x-coordinate: $x_{\text{avg}} = \frac{4 \cdot (-2) + 1 \cdot 2}{4 + 1} = \frac{-8 + 2}{5} = \frac{-6}{5} = -1.2$

• For the y-coordinate: $y_{\text{avg}} = \frac{4 \cdot 1 + 1 \cdot 5}{4 + 1} = \frac{4 + 5}{5} = \frac{9}{5} = 1.8$

Thus, the coordinates of the weighted average point are: $P_{\text{avg}} = (-1.2, 1.8)$

Let me know if you need further details or have any questions!

Here are 5 related questions for further exploration:

1. How does changing the weights affect the position of the weighted average point?
2. What would happen if both points had equal weights?
3. Can you find the weighted average if one of the weights is negative?
4. How can we generalize the weighted average formula for more than two points?
5. How do the coordinates of the midpoint relate to the weighted average in this case?

Tip: When working with weighted averages, always check that the sum of the weights is non-zero to avoid division errors.