Question ID: #1029
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:
- (1) $90$
- (2) $93$
- (3) $97$
- (4) $83$
Solution:
The equation of the given line $L$ is:
$$x + by + c = 0$$
Transforming it into intercept form $\frac{x}{X} + \frac{y}{Y} = 1$:
$$x + by = -c$$
$$\frac{x}{-c} + \frac{y}{-c/b} = 1$$
The x-intercept is $-c$ and the y-intercept is $-\frac{c}{b}$.
Since the perpendicular drawn from the origin to the line $L$ makes an angle of $45^{\circ}$ with the positive x-axis, the slope of the perpendicular is $\tan(45^{\circ}) = 1$.
The slope of line $L$ will be:
$$m_L = -\frac{1}{\text{slope of perpendicular}} = -\frac{1}{1} = -1$$
From the line equation $x + by + c = 0$, the slope is $-\frac{1}{b}$:
$$-\frac{1}{b} = -1 \Rightarrow b = 1$$
The area of a triangle formed by a line with coordinate axes is $\frac{1}{2}|\text{x-intercept} \times \text{y-intercept}|$:
$$\text{Area} = \frac{1}{2} |-c| \left|-\frac{c}{b}\right| = 48$$
$$\frac{c^2}{2|b|} = 48$$
Substitute $b = 1$:
$$\frac{c^2}{2(1)} = 48$$
$$c^2 = 96$$
Calculate $b^2 + c^2$:
$$b^2 + c^2 = (1)^2 + 96 = 1 + 96 = 97$$
Ans. (3)
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