Let the area of the triangle formed by a straight line $L : x + by + c = 0$ with coordinate 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:

Step 1: Find intercepts and area

First, we convert the given equation of the line $L: x + by + c = 0$ into the intercept form $\frac{x}{a} + \frac{y}{d} = 1$. This allows us to identify the $x$ -intercept as $a = -c$ and the $y$ -intercept as $d = -c/b$. The area of the triangle formed by the line with the coordinate axes is given by half the product of the absolute values of the intercepts. We are given that this area is $48$ square units, so we set up the equation $\frac{c^2}{2|b|} = 48$.

Step 2: Use perpendicular distance and angle

The perpendicular distance from the origin to the line $x + by + c = 0$ is given by the formula $p = \frac{|c|}{\sqrt{1^2 + b^2}}$. The direction cosines of the normal to the line $Ax + By + C = 0$ are $\frac{A}{\sqrt{A^2+B^2}}$ and $\frac{B}{\sqrt{A^2+B^2}}$. Since the perpendicular from the origin makes an angle of $45^\circ$ with the positive $x$ -axis, its direction cosines are $\cos 45^\circ$ and $\sin 45^\circ$. Comparing these, we find that $\frac{1}{\sqrt{1+b^2}} = \cos 45^\circ = \frac{1}{\sqrt{2}}$. Solving this equation gives $1+b^2 = 2$, which implies $b^2 = 1$.

Step 3: Calculate c squared

Now that we have $b^2 = 1$, we can substitute this value back into the area equation we derived earlier: $\frac{c^2}{2|b|} = 48$. Since $b^2 = 1$, $|b| = \sqrt{1} = 1$. Substituting this into the equation gives $\frac{c^2}{2(1)} = 48$. Multiplying both sides by $2$ yields $c^2 = 96$.

Step 4: Find the value of b squared plus c squared

Finally, we need to find the value of $b^2 + c^2$. We have already calculated $b^2 = 1$ and $c^2 = 96$. Adding these two values together, we get $1 + 96 = 97$.

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