1. Determine the vertices and area of the initial triangle:
The boundary lines are given as $$2x + y = 4$$, the positive x-axis ($$y = 0$$), and the positive y-axis ($$x = 0$$).
- To find the x-intercept, we set $$y = 0 \implies 2x = 4 \implies x = 2$$. This gives vertex $$(2, 0)$$.
- To find the y-intercept, we set $$x = 0 \implies y = 4$$. This gives vertex $$(0, 4)$$.
The third vertex sits at the origin $$(0, 0)$$.
The total area of this right-angled triangle is:
$$\text{Total Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 4 = 4$$
2. Set up the intercepts of the bisecting line:
The line cuts the area exactly in half, meaning the smaller triangle it forms with the origin must have an area of:
$$\text{New Area} = \frac{4}{2} = 2$$
Let the area-bisecting line make an x-intercept at $$x_0$$ and a y-intercept at $$y_0$$. The area equation gives:
$$\frac{1}{2} \cdot x_0 \cdot y_0 = 2 \implies x_0 y_0 = 4$$
3. Use the point constraint to find the intercepts:
The equation of the line written in its standard intercept form is:
$$\frac{x}{x_0} + \frac{y}{y_0} = 1$$
Since the line is explicitly stated to pass through the point $$(1, 1)$$, we substitute these coordinates into the equation:
$$\frac{1}{x_0} + \frac{1}{y_0} = 1 \implies \frac{x_0 + y_0}{x_0 y_0} = 1$$
Substitute our known area product value $$x_0 y_0 = 4$$ into this relation:
$$\frac{x_0 + y_0}{4} = 1 \implies x_0 + y_0 = 4$$
4. Calculate the required absolute sum:
The problem asks for the absolute value of the sum of its intercepts on the coordinate axes, which is:
$$|x_0 + y_0| = |4| = 4$$
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