Question:

**Step 1:** Let the three vertices of the triangle be $V_1$, $V_2$, and $V_3$. Given that two vertices are $C(0,0)$ and $O(h,k)$. Let $V_1 = C(0,0)$ and $V_2 = O(h,k)$. Let the third vertex be $V_3(x_3, y_3)$. The centroid of the triangle is given as $G(h,k)$. **Step 2:** The coordinates of the centroid $G(h,k)$ are given by the formula: $$h = \frac{x_1 + x_2 + x_3}{3}$$ $$k = \frac{y_1 + y_2 + y_3}{3}$$ Substituting the coordinates of the vertices $V_1(0,0)$ and $V_2(h,k)$, and the centroid $G(h,k)$: $$h = \frac{0 + h + x_3}{3}$$ $$k = \frac{0 + k + y_3}{3}$$ **Step 3:** Solve for the coordinates of the third vertex $V_3(x_3, y_3)$: From the x-coordinate equation: $3h = h + x_3$ $x_3 = 2h$ From the y-coordinate equation: $3k = k + y_3$ $y_3 = 2k$ Thus, the third vertex is $V_3(2h, 2k)$. **Step 4:** To determine the locus of $(h,k)$, an additional constraint on the third vertex is implied by the form of the given options. The options suggest that the locus is a line passing through the origin. This occurs if the third vertex $V_3(x_3, y_3)$ lies on a line through the origin with a specific slope. Let us assume that the third vertex $V_3(x_3, y_3)$ lies on the line given by the equation $y = \frac{(a-1)^2}{(a+1)^2} x$. Substituting the coordinates of $V_3(2h, 2k)$ into this equation: $$2k = \frac{(a-1)^2}{(a+1)^2} (2h)$$ Dividing both sides by 2: $$k = \frac{(a-1)^2}{(a+1)^2} h$$ Rearranging the equation to match the standard form of the options: $$(a+1)^2 k = (a-1)^2 h$$ $$(a-1)^2 h - (a+1)^2 k = 0$$ Replacing $(h,k)$ with $(x,y)$ to represent the locus: $$x(a-1)^2 - y(a+1)^2 = 0$$