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 \( = \frac{1}{2} \left| \frac{c^2}{b} \right| = 48 \) \(\left| \frac{c^2}{b} \right| = 96 \) 
\(\Rightarrow -c = - \frac{c}{b} \) \(\Rightarrow b = 1 \quad \Rightarrow c^2 = 96 \) \(\Rightarrow b^2 + c^2 = 97 \)

Answer 2

We are given a straight line \( L: x + by + c = 0 \) that cuts the coordinate axes and encloses a triangle with area \( 48 \, \text{sq. units} \). The perpendicular from the origin to this line makes an angle of \( 45^\circ \) with the positive x-axis. We must find \( b^2 + c^2 \).

Concept Used:

The general equation of a line is \( Ax + By + C = 0 \). The perpendicular distance from the origin to this line is given by:

\[ p = \frac{|C|}{\sqrt{A^2 + B^2}} \]

The slope of the perpendicular from the origin is the same as the direction of the normal vector to the line, which is \( \mathbf{n} = (A, B) \). The angle \( \theta \) that the perpendicular makes with the x-axis satisfies:

\[ \tan \theta = \frac{B}{A} \]

Here, \( A = 1, B = b, C = c \).

Step-by-Step Solution:

Step 1: From the given information, since the perpendicular makes a \(45^\circ\) angle with the positive x-axis,

\[ \tan 45^\circ = \frac{b}{1} \Rightarrow b = 1. \]

Step 2: Find the intercepts of the line on the coordinate axes.

For the x-intercept, set \( y = 0 \):

\[ x + c = 0 \Rightarrow x = -c. \]

For the y-intercept, set \( x = 0 \):

\[ b y + c = 0 \Rightarrow y = -\frac{c}{b}. \]

Hence, the intercepts are \( (-c, 0) \) and \( (0, -\tfrac{c}{b}) \).

Step 3: The area of the triangle formed by the line with the coordinate axes is given by:

\[ \text{Area} = \frac{1}{2} \times |x\text{-intercept}| \times |y\text{-intercept}| \] \[ 48 = \frac{1}{2} \times |c| \times \left|\frac{c}{b}\right| \] \[ 48 = \frac{c^2}{2|b|} \]

Step 4: Substitute \( b = 1 \):

\[ 48 = \frac{c^2}{2(1)} \Rightarrow c^2 = 96. \]

Step 5: Calculate \( b^2 + c^2 \):

\[ b^2 + c^2 = 1^2 + 96 = 97. \]

Final Answer: \( \boxed{97} \)