Question

The quadratic equation whose roots are the \( x \) and \( y \) intercepts of the line passing through \( (1, 1) \) and making a triangle of area \( A \) with the co-ordinate axes is

• A. \( x^2 + Ax + 2A = 0 \)
• B. \( x^2 - Ax + 2A = 0 \)
• C. \( x^2 - Ax - 2A = 0 \)
• D. None of these

Hint:

The area of a triangle formed by intercepts on the coordinate axes is \( \frac{1}{2} \times \text{base} \times \text{height} \).

Answer 1

The Correct Option is B

Step 1: Analyze the line equation.
The line passing through \( (1,1) \) has the general equation \( x + y = 2 \), which intersects the axes at \( x = 2 \) and \( y = 2 \).
Step 2: Use the area formula.
The area of the triangle formed by the intercepts is given by: \[ A = \frac{1}{2} \times 2 \times 2 = 2 \] Thus, the quadratic equation becomes \( x^2 - Ax + 2A = 0 \). Final Answer: \[ \boxed{x^2 - Ax + 2A = 0} \]