1. Set up coordinates

Take the $1.0\,\text{kg}$ mass as the origin.

From the usual right triangle arrangement shown:

• $1.0\,\text{kg}$ at $(0,0)$
• $1.5\,\text{kg}$ at $(4,0)$ cm
• $2.5\,\text{kg}$ at $(0,3)$ cm

This matches the $3$-$4$-$5$ right triangle.

Total mass: $M = 1.0+1.5+2.5 = 5.0\,\text{kg}$

2. Use center of mass formula

The coordinates of center of mass are $x_{\text{cm}} = \frac{\sum m_i x_i}{\sum m_i}, \qquad y_{\text{cm}} = \frac{\sum m_i y_i}{\sum m_i}$

3. Compute $x$-coordinate

$x_{\text{cm}} = \frac{(1.0)(0) + (1.5)(4) + (2.5)(0)}{5.0}$ $x_{\text{cm}} = \frac{6}{5} = 1.2\,\text{cm}$

4. Compute $y$-coordinate

$y_{\text{cm}} = \frac{(1.0)(0) + (1.5)(0) + (2.5)(3)}{5.0}$ $y_{\text{cm}} = \frac{7.5}{5} = 1.5\,\text{cm}$

5. Interpretation

So the center of mass is at $\boxed{1.2\,\text{cm right and }1.5\,\text{cm above the }1\,\text{kg mass}}$

6. Compare with options

Given options:

• A: $(2.0, 0.9)$
• B: $(0.9, 2.0)$
• C: $(0.6, 2.0)$
• D: $(1.5, 1.2)$

None matches exactly. The closest is option D if coordinates were interchanged approximately, but it is still not correct.

Hence, based on the standard interpretation of the figure, none of the listed options is correct.

If the masses at the other two vertices were swapped:

• $2.5\,\text{kg}$ at $(4,0)$
• $1.5\,\text{kg}$ at $(0,3)$

then $x_{\text{cm}} = \frac{(2.5)(4)}{5} = 2.0\,\text{cm}, \qquad y_{\text{cm}} = \frac{(1.5)(3)}{5} = 0.9\,\text{cm}$ which gives option A.

So the stored answer $B$ does not agree with the physics calculation under either natural placement.