1. Find the fixed point of intersection $$P$$:
We solve the linear system of equations to isolate our fixed pivot point:
$$x + 2y = 3$$
$$2x - y = 1 \implies y = 2x - 1$$
Substitute the expression for $$y$$ directly into the first equation:
$$x + 2(2x - 1) = 3$$
$$x + 4x - 2 = 3$$
$$5x = 5 \implies x = 1$$
$$y = 2(1) - 1 = 1$$
So, the fixed point of intersection is $$P(1, 1)$$.
2. Apply the maximum distance geometric principle:
For a family of lines pivoting through a fixed point $$P$$, the perpendicular distance from the origin $$O(0,0)$$ to the line reaches its absolute maximum length when the line itself is drawn perfectly perpendicular to the position segment line $$OP$$.
The slope of the straight segment line $$OP$$ connecting $$(0,0)$$ and $$(1,1)$$ is:
$$m_{OP} = \frac{1 - 0}{1 - 0} = 1$$
Since our target variable line must stand perpendicular to $$OP$$, its matching slope $$m$$ is:
$$m \times m_{OP} = -1 \implies m = -1$$
3. Formulate the equation of the line:
The line has a slope of $$-1$$ and passes through the coordinate point $$P(1,1)$$. Using the point-slope formula:
$$y - 1 = -1(x - 1)$$
$$y - 1 = -x + 1$$
$$x + y - 2 = 0$$
4. Compare with the standard form and evaluate:
Comparing $$x + y - 2 = 0$$ to the template $$ax + by + c = 0$$, we extract the parameters:
$$a = 1$$
$$b = 1$$
$$c = -2$$
Substitute these coefficient values into our target expression:
$$a^2 + b^2 - c^2 = (1)^2 + (1)^2 - (-2)^2 = 1 + 1 - 4 = -2$$
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