Solution:

We are tasked with finding the value of $x$ in a triangle geometry problem where a line divides the triangle into two equal-area regions. Let’s solve this step-by-step.

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Step 1: Coordinates of the centroid

The centroid $G$ of triangle $\triangle ABC$ is given by the formula: $G\left( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \right),$ where the vertices are $A(0, 0), B(1, 0), C(x, y)$.

Substituting the coordinates: $G\left( \frac{0+1+x}{3}, \frac{0+0+y}{3} \right) = \left( \frac{1+x}{3}, \frac{y}{3} \right).$

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Step 2: Area of $\triangle ABC$

The area of $\triangle ABC$ is: $\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|.$

Substituting $A(0, 0), B(1, 0), C(x, y)$: $\text{Area of } \triangle ABC = \frac{1}{2} \left| 0(0-y) + 1(y-0) + x(0-0) \right| = \frac{1}{2} |y|.$

Thus: $\text{Area of } \triangle ABC = \frac{y}{2}.$

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Step 3: The line $y = mx$ passing through $G$

The line $y = mx$ passes through the centroid $G\left( \frac{1+x}{3}, \frac{y}{3} \right)$. Substituting this point into the line equation: $\frac{y}{3} = m \cdot \frac{1+x}{3}.$

Simplify: $y = m(1+x).$

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Step 4: Ratio of areas

The problem states that the line $y = mx$ divides $\triangle ABC$ into two triangles of equal area. This means: $\text{Ratio of total area of } \triangle ABC \text{ to the area of the triangle formed by } y=mx \text{ and the x-axis} = \frac{-9}{4}.$

Let’s compute the area of the triangle formed by $y = mx$ and the x-axis. This triangle lies below the line $y = mx$ and has its base along the x-axis from $x = 0$ to $x = \frac{y}{m}$, and height $y$.

The area is: $\text{Area} = \frac{1}{2} \cdot \frac{y}{m} \cdot y = \frac{y^2}{2m}.$

Now, the ratio of areas is: $\frac{\text{Area of } \triangle ABC}{\text{Area of triangle formed by } y=mx \text{ and x-axis}} = \frac{\frac{y}{2}}{\frac{y^2}{2m}} = \frac{m}{y}.$

The problem states this ratio is $-\frac{9}{4}$: $\frac{m}{y} = -\frac{9}{4}.$

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Step 5: Solve for $x$

From the above equation: $m = -\frac{9y}{4}.$

Substituting $m = \frac{y}{1+x}$ (from Step 3): $\frac{y}{1+x} = -\frac{9y}{4}.$

Cancel $y$ (since $y > 0$): $\frac{1}{1+x} = -\frac{9}{4}.$

Cross-multiply: $4 = -9(1+x).$

Simplify: $4 = -9 - 9x.$

$9x = -13 \quad \Rightarrow \quad x = -\frac{13}{9}.$

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Final Answer:

$x = -\frac{13}{9}.$

Would you like additional clarification on the steps? Here are five related questions for deeper understanding:

1. How is the centroid of a triangle calculated?
2. Why does the line $y = mx$ divide the triangle into two equal areas?
3. How do we compute the area of a triangle formed by a line and the x-axis?
4. What is the significance of the ratio $-\frac{9}{4}$?
5. Can this method be generalized for other triangles?

Tip: Always double-check your work for errors in substituting equations or solving fractions.