Slope of a Line

Rearrange the family of lines by grouping terms with $$\lambda$$:

$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$

This is in the form $$L_1 + \lambda L_2 = 0$$. To find the fixed point $$P$$, solve the system:

$$x + 2y = 5$$

$$3x + 7y = 17$$

Multiplying the first equation by 3 and subtracting gives $$y = 2$$, which means $$x = 1$$. So, $$P = (1, 2)$$.

The line from this family that is farthest from the origin is the line perpendicular to the segment $$OP$$ (where $$O$$ is the origin $$(0,0)$$).

Slope of $$OP = \frac{2 - 0}{1 - 0} = 2$$. Therefore, the slope of line $$L$$ is $$-\frac{1}{2}$$.

Equation of line $$L$$ passing through $$P(1, 2)$$:

$$y - 2 = -\frac{1}{2}(x - 1) \implies x + 2y - 5 = 0$$

$$d = \frac{|3 + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$$

$$d^2 = \frac{100}{5} = 20$$

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:

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

The coordinates of the given points are $$A(2,5)$$ and $$B(6,1)$$.

Step 1: Equation of the perpendicular bisector $$L$$.
• Mid-point of $$AB$$: $$M\left(\dfrac{2+6}{2},\dfrac{5+1}{2}\right)=(4,3)$$.
• Slope of $$AB$$: $$m_{AB}=\dfrac{1-5}{6-2}=\dfrac{-4}{4}=-1$$.
• Hence slope of the perpendicular bisector is the negative reciprocal, i.e. $$+1$$.
Therefore $$L$$ is $$y-3 = 1(x-4) \;\Rightarrow\; y = x-1\;.$$

Step 2: Area condition for $$\triangle PAB$$.
Let $$P(x,y)$$ lie on $$L$$. The area of a triangle formed by points $$P(x,y),\;A(2,5),\;B(6,1)$$ is

$$\text{Area}=\dfrac12 \left|\begin{vmatrix} 2 & 5 & 1\\ 6 & 1 & 1\\ x & y & 1 \end{vmatrix}\right| \;.$$

Expanding the determinant,

$$\begin{aligned} D &= 2(1-y) - 5(6-x) + 1(6y - x)\\ &= 2 - 2y - 30 + 5x + 6y - x\\ &= -28 + 4x + 4y. \end{aligned}$$

The area is given to be $$10$$ square units, hence

$$\dfrac12|D| = 10 \;\Longrightarrow\; |-28 + 4x + 4y| = 20.$$

Dividing by 4,

$$|-7 + x + y| = 5 \;.\quad -(1)$$

Step 3: Impose the line condition $$y = x - 1$$.
Substitute $$y = x-1$$ into $$(1)$$:

$$\bigl| -7 + (x + x - 1) \bigr| = 5 \;\Longrightarrow\; |\,2x - 8\,| = 5.$$

Step 4: Solve for $$x$$ (two cases).

Case 1: $$2x - 8 =\; 5 \;\Rightarrow\; 2x = 13 \;\Rightarrow\; x = 6.5.$$ Then $$y = x - 1 = 5.5.$$ So $$P_1(6.5,\,5.5).$$

Case 2: $$2x - 8 = -5 \;\Rightarrow\; 2x = 3 \;\Rightarrow\; x = 1.5.$$ Then $$y = x - 1 = 0.5.$$ So $$P_2(1.5,\,0.5).$$

Step 5: Required sum of all the x and y coordinates.
Add the x- and y-coordinates of both admissible points:

$$\begin{aligned} \text{Total} &= (6.5 + 5.5) + (1.5 + 0.5) \\ &= 12 + 2 = 14. \end{aligned}$$

Hence the sum equals $$14$$.

Option C which is: $$14$$