Let the area of the triangle formed by a straight Line L: $\mathrm{x}+\mathrm{by}+\mathrm{c}=0$ 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 $\mathrm{b}^2+\mathrm{c}^2$…

Question

Let the area of the triangle formed by a straight Line L: $\mathrm{x}+\mathrm{by}+\mathrm{c}=0$ 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 $\mathrm{b}^2+\mathrm{c}^2$ is:

Aicharya · Accepted Answer

$\frac{x}{-c}+\frac{y}{-c / b}=1$

$\begin{aligned} & \therefore \text { area of triangle }=\frac{1}{2}\left|\frac{c^2}{b}\right|=48 \\ & \left|\frac{c^2}{b}\right|=96 \\ & \because-c=-\frac{c}{b} \\ & \Rightarrow b=1 \quad \therefore c^2=96 \\ & \therefore b^2+c^2=97\end{aligned}$ ^

Let the area of the triangle formed by a straight Line L: $\mathrm{x}+\mathrm{by}+\mathrm{c}=0$ with…

Solution