Question

Let A(a, 0), B(b, 2b + 1) and C(0, b), b \(\neq\) 0, |b| \(\neq\) 1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :

• A. \(\frac{2b}{b+1}\)
• B. \(-\frac{2b}{b+1}\)
• C. \(\frac{2b^2}{b+1}\)
• D. \(-\frac{2b^2}{b+1}\)

Hint:

When dealing with an absolute value equation like |X| = k (where k>0), always remember to solve for two separate cases: X = k and X = -k. This often leads to multiple possible solutions for the variables involved.

Answer 1

The Correct Option is D

Step 1: Use the formula for the area of a triangle.
The area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) is given by: Area = \( \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| \) Given vertices are A(a, 0), B(b, 2b+1), and C(0, b). The area is 1. \[ 1 = \frac{1}{2} |a((2b+1) - b) + b(b - 0) + 0(0 - (2b+1))| \] \[ 2 = |a(b+1) + b^2| \] Step 2: Solve the equation for a.
The absolute value equation gives two possibilities: Case 1: \(a(b+1) + b^2 = 2\) \[ a(b+1) = 2 - b^2 \] \[ a_1 = \frac{2 - b^2}{b+1} \] Case 2: \(a(b+1) + b^2 = -2\) \[ a(b+1) = -2 - b^2 \] \[ a_2 = \frac{-2 - b^2}{b+1} \] These are the two possible values of a. Step 3: Find the sum of all possible values of a.
Sum = \(a_1 + a_2\) \[ \text{Sum} = \frac{2 - b^2}{b+1} + \frac{-2 - b^2}{b+1} \] \[ \text{Sum} = \frac{(2 - b^2) + (-2 - b^2)}{b+1} = \frac{2 - b^2 - 2 - b^2}{b+1} = \frac{-2b^2}{b+1} \]