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(C) The equation of the line is $x + by + c = 0$,which can be written as $\frac{x}{-c} + \frac{y}{-c/b} = 1$. The intercepts are $a = -c$ and $b' = -c

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$97$

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(C) The equation of the line is $x + by + c = 0$,which can be written as $\frac{x}{-c} + \frac{y}{-c/b} = 1$. The intercepts are $a = -c$ and $b' = -c/b$. The area of the triangle is $\frac{1}{2} |a \cdot b'| = \frac{1}{2} |(-c) \cdot (-c/b)| = \frac{1}{2} |\frac{c^2}{b}| = 48$. Thus,$|\frac{c^2}{b}| = 96$. The perpendicular from the origin to the line $L$ makes an angle of $45^{\circ}$ with the positive $x$-axis. The slope of this perpendicular is $	an(45^{\circ}) = 1$. The slope of the line $L$ is $-1/b$. Since the product of slopes of perpendicular lines is $-1$,we have $(1) \cdot (-1/b) = -1$,which gives $b = 1$. Substituting $b = 1$ into $|\frac{c^2}{b}| = 96$,we get $|c^2| = 96$,so $c^2 = 96$. Therefore,$b^2 + c^2 = 1^2 + 96 = 97$.

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

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