If straight lines ax + by + p = 0 and x cos a | vedclass

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(B) The lines $ax + by + p = 0$ and $x \cos \alpha + y \sin \alpha - p = 0$ are inclined at an angle $\pi / 4$.Therefore,$	an(\pi / 4) = \left| \frac

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$2$

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(B) The lines $ax + by + p = 0$ and $x \cos \alpha + y \sin \alpha - p = 0$ are inclined at an angle $\pi / 4$.Therefore,$	an(\pi / 4) = \left| \frac{(-a/b) - (-\cos \alpha / \sin \alpha)}{1 + (-a/b)(-\cos \alpha / \sin \alpha)} ight| = 1$.This simplifies to $|-a \sin \alpha + b \cos \alpha| = |b \sin \alpha + a \cos \alpha|$.Squaring both sides,$a^2 \sin^2 \alpha + b^2 \cos^2 \alpha - 2ab \sin \alpha \cos \alpha = b^2 \sin^2 \alpha + a^2 \cos^2 \alpha + 2ab \sin \alpha \cos \alpha$.Rearranging gives $(a^2 - b^2)(\sin^2 \alpha - \cos^2 \alpha) = 4ab \sin \alpha \cos \alpha$,which leads to $(a^2 - b^2)(-\cos 2\alpha) = 2ab \sin 2\alpha$.Since the lines are concurrent with $x \sin \alpha - y \cos \alpha = 0$,the determinant of the coefficients is zero:$\begin{vmatrix} a & b & p \ \cos \alpha & \sin \alpha & -p \ \sin \alpha & -\cos \alpha & 0 \end{vmatrix} = 0$.Expanding the determinant: $a(-p \cos \alpha) - b(p \sin \alpha) + p(-\cos^2 \alpha - \sin^2 \alpha) = 0$.$-ap \cos \alpha - bp \sin \alpha - p = 0 \implies a \cos \alpha + b \sin \alpha = -1$.Squaring this: $a^2 \cos^2 \alpha + b^2 \sin^2 \alpha + 2ab \sin \alpha \cos \alpha = 1$.Using the concurrency and angle conditions,we find $a^2 + b^2 = 2$.

If straight lines $ax + by + p = 0$ and $x \cos \alpha + y \sin \alpha - p = 0$ include an angle $\pi / 4$ between them and meet the straight line $x \sin \alpha - y \cos \alpha = 0$ in the same point,then the value of $a^2 + b^2$ is equal to

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