Key Concepts and Formulas

• Area of a Triangle using Determinants: The area of a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ can be calculated using the formula: $\text{Area} = \frac{1}{2} \left| \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} \right|$ The absolute value is crucial because area is always a non-negative quantity.
• Property of Absolute Value Equations: If $|X| = K$ where $K > 0$, then $X = K$ or $X = -K$. This property is vital when solving for unknown variables within an absolute value expression.
• Determinant Expansion: A $3 \times 3$ determinant can be expanded along any row or column. For a matrix $\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$, expansion along the first row is $a(ei-fh) - b(di-fg) + c(dh-eg)$. Choosing a row or column with zero elements can simplify calculations.

Step-by-Step Solution

Step 1: Set up the Area Equation using Determinants.

We are given the coordinates of the three vertices: $A(a, 0)$, $B(b, 2b + 1)$, and $C(0, b)$. The area of triangle ABC is given as 1 square unit. We substitute these coordinates into the area formula:

$1 = \frac{1}{2} \left| \det \begin{pmatrix} a & 0 & 1 \\ b & 2b+1 & 1 \\ 0 & b & 1 \end{pmatrix} \right|$

To simplify, we multiply both sides of the equation by 2. This isolates the absolute value of the determinant, making it easier to proceed:

$2 = \left| \det \begin{pmatrix} a & 0 & 1 \\ b & 2b+1 & 1 \\ 0 & b & 1 \end{pmatrix} \right|$

Let $D$ represent the determinant. So, we have $|D| = 2$.

Step 2: Evaluate the Determinant.

Now, we calculate the value of the determinant $D$: $D = \det \begin{pmatrix} a & 0 & 1 \\ b & 2b+1 & 1 \\ 0 & b & 1 \end{pmatrix}$

To make the calculation efficient, we expand the determinant along the first row ($R_1$) because it contains a zero element. This reduces the number of terms we need to compute.

Expanding along $R_1$: $D = a \cdot \begin{vmatrix} 2b+1 & 1 \\ b & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} b & 1 \\ 0 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} b & 2b+1 \\ 0 & b \end{vmatrix}$

Next, we evaluate the $2 \times 2$ determinants (minors):

• For the first term: $a \cdot ((2b+1)(1) - (1)(b)) = a \cdot (2b+1 - b) = a(b+1)$
• The second term is $0 \cdot (\text{any value})$, which simplifies to $0$.
• For the third term: $1 \cdot ((b)(b) - (2b+1)(0)) = 1 \cdot (b^2 - 0) = b^2$

Combining these results, the value of the determinant is: $D = a(b+1) + 0 + b^2$ $D = a(b+1) + b^2$

Step 3: Formulate Two Cases for the Determinant and Solve for 'a'.

From Step 1, we have $|D| = 2$. Using the property of absolute value equations, this implies that the determinant $D$ can be either $2$ or $-2$. We must consider both possibilities to find all possible values of $a$.

Case 1: The determinant is equal to 2 Set the expression for $D$ equal to $2$: $a(b+1) + b^2 = 2$ To solve for $a$, we isolate the term containing $a$: $a(b+1) = 2 - b^2$ We are given that $b \ne 0$ and $|b| \ne 1$. The condition $|b| \ne 1$ implies $b \ne 1$ and $b \ne -1$. Therefore, $b+1 \ne 0$, which means we can safely divide by $(b+1)$: $a_1 = \frac{2 - b^2}{b+1}$

Case 2: The determinant is equal to -2 Set the expression for $D$ equal to $-2$: $a(b+1) + b^2 = -2$ Isolate the term containing $a$: $a(b+1) = -2 - b^2$ Again, since $b+1 \ne 0$, we can divide: $a_2 = \frac{-2 - b^2}{b+1}$

Step 4: Calculate the Sum of All Possible Values of 'a'.

The problem asks for the sum of all possible values of $a$, which are $a_1$ and $a_2$. $\text{Sum} = a_1 + a_2$ Substitute the expressions for $a_1$ and $a_2$: $\text{Sum} = \frac{2 - b^2}{b+1} + \frac{-2 - b^2}{b+1}$ Since both terms have the same denominator, we can add their numerators directly: $\text{Sum} = \frac{(2 - b^2) + (-2 - b^2)}{b+1}$ $\text{Sum} = \frac{2 - b^2 - 2 - b^2}{b+1}$ Combine like terms in the numerator: $\text{Sum} = \frac{-2b^2}{b+1}$

Common Mistakes & Tips

• Forgetting the Absolute Value: A common error is to forget the absolute value in the area formula, which would lead to only one case for the determinant and potentially miss valid solutions for 'a'.
• Ignoring Both Cases for Absolute Value: When solving $|D|=K$, it's crucial to consider both $D=K$ and $D=-K$. Failing to do so would result in finding only half of the possible values for 'a'.
• Determinant Calculation Errors: Be meticulous with signs and calculations when expanding determinants, especially with algebraic expressions involving variables.
• Checking Denominator Conditions: Always ensure that any division by an expression involving a variable is valid (i.e., the denominator is not zero). The problem statement ($|b| \ne 1$) explicitly ensures $b+1 \ne 0$.

Summary

This problem required us to utilize the determinant formula for the area of a triangle. After setting up the equation with the given area and vertex coordinates, we evaluated the $3 \times 3$ determinant. The critical step involved recognizing that the absolute value of the determinant being 2 leads to two distinct cases for the determinant's value: 2 and -2. We solved for 'a' in each case, yielding two possible values ($a_1$ and $a_2$). Finally, summing these two values provided the required answer, demonstrating the importance of thoroughly considering all mathematical possibilities arising from absolute value equations.

The final answer is $\boxed{\frac{-2b^2}{b+1}}$ which corresponds to option (A).