Condition for Perpendicular Lines

1. Find the original vertex $$V$$

Let the coordinates of $$V$$ be $$(x_1, y_1)$$. Its reflection across the line $$2x + 3y - 5 = 0$$ is given as $$V'(-1, -2)$$.

Using the standard line reflection formula: $$\frac{x_1 - x_2}{a} = \frac{y_1 - y_2}{b} = -2 \frac{ax_2 + by_2 + c}{a^2 + b^2}$$

Substituting our values ($$x_2 = -1$$, $$y_2 = -2$$, $$a = 2$$, $$b = 3$$, $$c = -5$$):

$$\frac{x_1 - (-1)}{2} = \frac{y_1 - (-2)}{3} = -2 \frac{2(-1) + 3(-2) - 5}{2^2 + 3^2}$$

$$\frac{x_1 + 1}{2} = \frac{y_1 + 2}{3} = -2 \left(\frac{-2 - 6 - 5}{13}\right) = -2 \left(\frac{-13}{13}\right) = 2$$

Now, solve for $$x_1$$ and $$y_1$$:

$$\frac{x_1 + 1}{2} = 2 \implies x_1 + 1 = 4 \implies x_1 = 3$$

$$\frac{y_1 + 2}{3} = 2 \implies y_1 + 2 = 6 \implies y_1 = 4$$

Thus, the vertex opposite to the hypotenuse is $$V(3, 4)$$.

2. Find the slopes of the perpendicular sides

The given hypotenuse line is $$4x + 3y = 5$$, which has a slope of $$m = -\frac{4}{3}$$.

Since the triangle is a right-angled isosceles triangle, both perpendicular sides intersect at $$V$$ and make an angle of $$45^\circ$$ with the hypotenuse. Let $$m_s$$ be the slope of a side:

$$\tan(45^\circ) = \left\vert \frac{m_s - \left(-\frac{4}{3}\right)}{1 + m_s\left(-\frac{4}{3}\right)} \right\vert \implies 1 = \left\vert \frac{3m_s + 4}{3 - 4m_s} \right\vert$$

This gives two cases:

Case 1:

$$3m_s + 4 = 3 - 4m_s \implies 7m_s = -1 \implies m_s = -\frac{1}{7}$$

Case 2:

$$3m_s + 4 = -(3 - 4m_s) \implies 3m_s + 4 = -3 + 4m_s \implies m_s = 7$$

(Note that the product of the two slopes is $$(-\frac{1}{7}) \times 7 = -1$$, confirming they are perpendicular).

3. Write the equations of the sides

Both lines pass through the calculated vertex $$V(3, 4)$$: 

Side 1 (with slope $$m_s = 7$$): $$y - 4 = 7(x - 3) \implies y - 4 = 7x - 21 \implies 7x - y = 17$$

Side 2 (with slope $$m_s = -\frac{1}{7}$$): $$y - 4 = -\frac{1}{7}(x - 3) \implies 7y - 28 = -x + 3 \implies x + 7y = 31$$

Correct Answer is  (C) $$7x - y = 17$$ and $$x + 7y = 31$$

In an isosceles triangle, the angle bisector of the two equal sides is perpendicular to the third side (the base). Therefore, the slope of the third side ($$m$$) will be perpendicular to the slopes of the angle bisectors.
Let the given lines be $$L_1: -x + 2y - 4 = 0$$ and $$L_2: x + y - 4 = 0$$.
The equations of the angle bisectors are given by:
$$\frac{-x + 2y - 4}{\sqrt{(-1)^2 + 2^2}} = \pm \frac{x + y - 4}{\sqrt{1^2 + 1^2}}$$
$$\frac{-x + 2y - 4}{\sqrt{5}} = \pm \frac{x + y - 4}{\sqrt{2}}$$
Cross-multiplying and rearranging gives two bisector lines with slopes $$m_{b1}$$ and $$m_{b2}$$.
Since the third side is perpendicular to these bisectors, calculating its possible slopes yields:
$$m = 3 + \sqrt{10} \quad \text{and} \quad m = 3 - \sqrt{10}$$
Sum of possible values:
$$\text{Sum} = (3 + \sqrt{10}) + (3 - \sqrt{10}) = 6$$

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:

$$a^2 + b^2 - c^2 = (1)^2 + (1)^2 - (-2)^2 = 1 + 1 - 4 = -2$$