Key Concepts and Formulas

• Area of a Triangle using Determinants: The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by: $\text{Area} = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$
• Equation of a Line: Given two points $(x_1, y_1)$ and $(x_2, y_2)$, the equation of the line passing through them can be found using the slope-point form or the determinant form.
• Position of a Point with Respect to a Line: A point $(x_0, y_0)$ is above the line $ax + by + c = 0$ if $ax_0 + by_0 + c$ has the same sign as $a x + by + c$ when evaluated for a point known to be above the line.

Step-by-Step Solution

Step 1: Define the coordinates and set up the area equation.

Let the variable point P be $(x, y)$. The given points are A$(-1, 1)$ and B$(2, 3)$. The area of $\triangle \mathrm{PAB}$ is given as 10. We will use the determinant formula for the area of a triangle. $\frac{1}{2} \left| \begin{vmatrix} x & y & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} \right| = 10$

Step 2: Expand the determinant.

We expand the determinant along the first row: $\begin{vmatrix} x & y & 1 \\ -1 & 1 & 1 \\ 2 & 3 & 1 \end{vmatrix} = x(1 - 3) - y(-1 - 2) + 1(-3 - 2) = -2x + 3y - 5$ So, the area equation becomes: $\frac{1}{2} |-2x + 3y - 5| = 10$ $|-2x + 3y - 5| = 20$

Step 3: Determine the sign of the expression inside the absolute value using the "P above line AB" condition.

The problem states that P is above the line AB. We need to determine the sign of $-2x + 3y - 5$. First, we find the equation of line AB. The slope of AB is $m = \frac{3 - 1}{2 - (-1)} = \frac{2}{3}$. Using the point-slope form with A$(-1, 1)$: $y - 1 = \frac{2}{3}(x + 1)$ $3y - 3 = 2x + 2$ $2x - 3y + 5 = 0$ Now, consider the expression $2x - 3y + 5$. If a point P$(x, y)$ is above the line AB, then $2x - 3y + 5 < 0$. To see why, test a point above the line, say $(0, 2)$. Then $2(0) - 3(2) + 5 = -1 < 0$.

Since $2x - 3y + 5 < 0$ for points above the line, $-(2x - 3y + 5) > 0$. Therefore, $-2x + 3y - 5 > 0$. This justifies removing the absolute value and keeping the expression as is: $-2x + 3y - 5 = 20$

Step 4: Simplify the equation and find a and b.

Rearrange the equation: $-2x + 3y = 25$ We want to express the locus in the form $ax + by = 15$. Multiply both sides by $\frac{3}{5}$: $\frac{3}{5}(-2x + 3y) = \frac{3}{5}(25)$ $-\frac{6}{5}x + \frac{9}{5}y = 15$ So, $a = -\frac{6}{5}$ and $b = \frac{9}{5}$.

Step 5: Calculate 5a + 2b.

Now, we calculate $5a + 2b$: $5a + 2b = 5\left(-\frac{6}{5}\right) + 2\left(\frac{9}{5}\right) = -6 + \frac{18}{5} = \frac{-30 + 18}{5} = -\frac{12}{5}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding the determinant and substituting into the equation of the line.
• Absolute Value: Remember that the absolute value results in two possible equations. The condition "P above line AB" is crucial for determining the correct sign.
• Scaling: Don't forget to scale the equation to match the required form before extracting the values of a and b.

Summary

We found the locus of point P by using the determinant formula for the area of a triangle and the condition that P lies above the line AB. We expanded the determinant, used the given area to form an equation involving an absolute value. By analyzing the position of P relative to the line AB, we removed the absolute value and simplified the equation. Finally, we scaled the equation to match the required form and calculated $5a + 2b$.

The final answer is \boxed{-\frac{12}{5}}, which corresponds to option (A).