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The problem is asking for the coordinates where a third particle (1.41 kg) must be placed such that the center of mass of a three-particle system is at the coordinates (-0.409 m, -0.699 m).

### Given information:
- Particle 1: Mass \( m_1 = 1.14 \, \text{kg} \), coordinates \( (-1.55 \, \text{m}, 0.271 \, \text{m}) \).
- Particle 2: Mass \( m_2 = 3.48 \, \text{kg} \), coordinates \( (0.793 \, \text{m}, -0.207 \, \text{m}) \).
- Particle 3: Mass \( m_3 = 1.41 \, \text{kg} \), coordinates \( (x, y) \) (unknown, to be determined).
- Center of mass: \( (-0.409 \, \text{m}, -0.699 \, \text{m}) \).

### Formula for the center of mass:
The center of mass coordinates \((x_{cm}, y_{cm})\) are calculated as:

\[
x_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}
\]
\[
y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3}{m_1 + m_2 + m_3}
\]

### Step 1: Solve for \( x_3 \)
Substitute the known values into the \( x_{cm} \) equation:

\[
-0.409 = \frac{(1.14)(-1.55) + (3.48)(0.793) + (1.41)(x)}{1.14 + 3.48 + 1.41}
\]
Simplifying:
\[
-0.409 = \frac{-1.767 + 2.761 + 1.41x}{6.03}
\]
\[
-0.409 \times 6.03 = -1.767 + 2.761 + 1.41x
\]
\[
-2.46527 = 0.994 - 1.767 + 1.41x
\]
\[
-2.46527 = -0.773 + 1.41x
\]
\[
1.41x = -2.46527 + 0.773
\]
\[
1.41x = -1.69227
\]
\[
x = \frac{-1.69227}{1.41}
\]
\[
x \approx -1.2 \, \text{m}
\]

### Step 2: Solve for \( y_3 \)
Substitute the known values into the \( y_{cm} \) equation:

\[
-0.699 = \frac{(1.14)(0.271) + (3.48)(-0.207) + (1.41)(y)}{1.14 + 3.48 + 1.41}
\]
Simplifying:
\[
-0.699 = \frac{0.30894 - 0.72036 + 1.41y}{6.03}
\]
\[
-0.699 \times 6.03 = 0.30894 - 0.72036 + 1.41y
\]
\[
-4.21557 = -0.41142 + 1.41y
\]
\[
1.41y = -4.21557 + 0.41142
\]
\[
1.41y = -3.80415
\]
\[
y = \frac{-3.80415}{1.41}
\]
\[
y \approx -2.7 \, \text{m}
\]

### Final answer:
The coordinates of the third particle must be approximately \( (-1.2 \, \text{m}, -2.7 \, \text{m}) \).

Let me know if you'd like further details, or if you have any questions!

### Related Questions:
1. What is the importance of the center of mass in physical systems?
2. How would the center of mass change if the mass of the third particle increased?
3. What happens to the center of mass if the first and second particles are swapped?
4. Can the center of mass lie outside the physical boundaries of an object or system?
5. How does the center of mass affect the stability of an object?

**Tip:** The center of mass is often referred to as the "balance point" of a system, which can be helpful in understanding its motion.

A 1.14 kg particle has the xy coordinates (-1.55 m, 0.271 m), and a 3.48 kg particle has the xy coordinates (0.793 m, -0.207 m). Both are on a horizontal plane. At what (a) x and (b) y coordinates must you place a 1.41 kg particle such that the center of mass of the three-particle system has the coordinates (-0.409 m, -0.699 m)?

Learn how to calculate the center of mass for a three-particle system where two particles have known coordinates and masses, and you need to find the coordinates of a third particle. This detailed solution involves center of mass formulas and algebraic manipulation.

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