The problem asks to find the weighted average of the points $W$, $X$, and $Y$ on the number line, where:

• Point $W$ weighs twice as much as point $X$,
• Point $Y$ weighs three times as much as point $X$.

Let's begin by identifying the coordinates of the points on the number line:

• $W = -6$,
• $X = -3$,
• $Y = 1$.

Let the weight of point $X$ be $1$ (since the weight is relative), which means:

• The weight of $W$ is $2 \times 1 = 2$,
• The weight of $Y$ is $3 \times 1 = 3$.

The formula for the weighted average is:

$P = \frac{(W \times \text{weight of } W) + (X \times \text{weight of } X) + (Y \times \text{weight of } Y)}{\text{sum of the weights}}$

Substituting the values:

$P = \frac{(-6 \times 2) + (-3 \times 1) + (1 \times 3)}{2 + 1 + 3}$

Let me calculate the result.The coordinate of point $P$, representing the weighted average of points $W$, $X$, and $Y$, is $-2$.

Would you like more details on how this was calculated or have any further questions?

Here are five related questions for practice:

1. What would the coordinate of $P$ be if point $Y$ weighed 4 times as much as point $X$ instead of 3?
2. How would the result change if point $X$ had a different weight?
3. Can you calculate the weighted average of four points on a number line?
4. What happens if all the points have equal weights?
5. How does changing the number line scale affect the weighted average?

Tip: When calculating a weighted average, always remember that the sum of the weights is crucial in determining the balance point.