If {p_1}, {p_2} and {p_3} are the perpendicul | vedclass

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(B) The perpendicular distance $p$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

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$G.P.$

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(B) The perpendicular distance $p$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Here,the line is $x \cos \alpha + y \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha} = 0$,which can be written as $x \cos^2 \alpha + y \sin \alpha \cos \alpha + \sin^2 \alpha = 0$.For point $({m^2}, 2m)$,$p_1 = |m^2 \cos^2 \alpha + 2m \sin \alpha \cos \alpha + \sin^2 \alpha| = |(m \cos \alpha + \sin \alpha)^2| = (m \cos \alpha + \sin \alpha)^2$.Similarly,for point $(m'^2, 2m')$,$p_3 = (m' \cos \alpha + \sin \alpha)^2$.For point $(mm', m + m')$,$p_2 = |mm' \cos^2 \alpha + (m + m') \sin \alpha \cos \alpha + \sin^2 \alpha| = |(m \cos \alpha + \sin \alpha)(m' \cos \alpha + \sin \alpha)| = |m \cos \alpha + \sin \alpha| \cdot |m' \cos \alpha + \sin \alpha|$.Thus,$p_2 = \sqrt{p_1} \cdot \sqrt{p_3}$,which implies $p_2^2 = p_1 p_3$.Therefore,${p_1}, {p_2}, {p_3}$ are in $G.P.$

If ${p_1}, {p_2}$ and ${p_3}$ are the perpendicular distances from the points $({m^2}, 2m)$,$(mm', m + m')$ and $(m'^2, 2m')$ respectively to the line $x \cos \alpha + y \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha} = 0$,then ${p_1}, {p_2}$ and ${p_3}$ are in:

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