Let the area of the triangle formed by a straight Line L: x + by + c = 0$L: x + by + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L$L$ makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, then the value of b^2 + c^2$b^2 + c^2$ is:

### Related Formula textNormal form of a straight line: x cos alpha + y sin alpha = p$$\text{Normal form of a straight line: } x \cos \alpha + y \sin \alpha = p$$ textArea of right-angled triangle formed with axes: A = frac12 left| x_textintercept cdot y_textintercept right|$$\text{Area of right-angled triangle formed with axes: } A = \frac{1}{2} \left| x_{\text{intercept}} \cdot y_{\text{intercept}} \right|$$ ### Core Logic We write down the normal equation of the straight line using the given polar normal angle alpha = 45^circ$\alpha = 45^\circ$, find its intercept coordinates, and use the area constraint to solve for the coefficients. ### Step 1: Write down normal form The perpendicular drawn from the origin makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, so alpha = 45^circ$\alpha = 45^\circ$. The line equation is: x cos 45^circ + y sin 45^circ = p implies fracxsqrt2 + fracysqrt2 = p$$x \cos 45^\circ + y \sin 45^\circ = p \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$$ x + y = psqrt2 implies x + y - psqrt2 = 0$$x + y = p\sqrt{2} \implies x + y - p\sqrt{2} = 0$$ Comparing this with the given format x + by + c = 0$x + by + c = 0$, we find: b = 1 quad textand quad c = -psqrt2$$b = 1 \quad \text{and} \quad c = -p\sqrt{2}$$ ### Step 2: Solve for the parameters using the area constraint The line equation is x + y = psqrt2$x + y = p\sqrt{2}$. The intercepts are: - x_textintercept = psqrt2$x_{\text{intercept}} = p\sqrt{2}$ - y_textintercept = psqrt2$y_{\text{intercept}} = p\sqrt{2}$ The area of the right-angled triangle formed with the axes is: textArea = frac12 left| psqrt2 cdot psqrt2 right| = p^2$$\text{Area} = \frac{1}{2} \left| p\sqrt{2} \cdot p\sqrt{2} \right| = p^2$$ Since the area is given as 48 square units: p^2 = 48$p^2 = 48$ ### Step 3: Calculate the requested value We have: b^2 = 1^2 = 1$b^2 = 1^2 = 1$ c^2 = (-psqrt2)^2 = 2p^2 = 2(48) = 96$$c^2 = (-p\sqrt{2})^2 = 2p^2 = 2(48) = 96$$ Therefore, we find: b^2 + c^2 = 1 + 96 = 97$$b^2 + c^2 = 1 + 96 = 97$$ ### Pattern Recognition Normal equation coupling: Normal equations of lines xcosalpha+ysinalpha = p$x\cos\alpha+y\sin\alpha = p$ are extremely powerful when normal angles are specified. For alpha=45^circ$\alpha=45^\circ$, the coordinate intercepts are identical, making the area relation A=p^2$A=p^2$ exceptionally simple. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines