Question

Let the area of the triangle formed by a straight line $ L: x + by + c = 0 $ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line $ L $ makes an angle of $ 45^\circ $ with the positive x-axis, then the value of $ b^2 + c^2 $ is:

• A. 90
• B. 93
• C. 97
• D. 83

Hint:

In problems like these, the area of the triangle can be found using the formula \( \frac{1}{2} \times \text{base} \times \text{height} \), where base and height are the distances from the origin to the x-axis and y-axis, respectively. Also, remember to use trigonometric relations to find these distances.

Answer 1

The Correct Option is C

\(\frac{x}{-c} + \frac{y}{-c/b} = 1\)  
Area of triangle \( = \frac{1}{2} \left| \frac{c^2}{b} \right| = 48 \) \(\left| \frac{c^2}{b} \right| = 96 \) 
\(\Rightarrow -c = - \frac{c}{b} \) \(\Rightarrow b = 1 \quad \Rightarrow c^2 = 96 \) \(\Rightarrow b^2 + c^2 = 97 \)