Area of Triangle from Determinant

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: $$14$$