Let the area of the triangle formed by a straight line L:x+by+c=0 with co-ordinate axes be…

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 :

Accepted Answer

$$\begin{aligned}
& L: x+b y+c=0 \
& \because \frac{1}{2}\left|(-c) \cdot\left(\frac{-c}{b}\right)\right|=48 \
& \therefore\left|\frac{c^2}{b}\right|=96\quad\text{..... (i)}
\end{aligned}$$

Slope of line $L=-\frac{1}{b}$

$\therefore$ Slope of line perpendicular to $L$ is $b$.

$$\begin{aligned}
& \therefore b=1 \
& \therefore c^2=96 \
& \therefore b^2+c^2=97
\end{aligned}$$

Solutions · 1