Question

If $p_1,p_2,p_3$ are length of perpendiculars from points $(m^2,2m),(mm-m+m)$ and $(m^{+2},2m)$ to the line $x\cos \alpha +y\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }=0$ then $p_1,p_2,p_3$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. $A.G.P$

Answer 1

The correct option is B $G.P.$

$\begin{array}{c}\Rightarrow p_1=\frac{\left|m^2\cos \alpha +2m\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }-0\right|}{\sqrt{{\sin }^2\lambda +{\cos }^2\lambda }}\\ \Rightarrow p_1=\left|m^2\cos \alpha +2m\sin \alpha +\tan \alpha +\sin \alpha \right|\\ \Rightarrow p_2=\frac{\left|\left(m^2-m\right)\cos \alpha +m\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }\right|}{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\\ \Rightarrow p_2=\left|m^2\cos \alpha +m\sin \alpha -m\cos \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }\right|\\ \Rightarrow p_3=\frac{\left|m^2\cos \alpha +2m\sin \alpha +\frac{{\sin }^2\alpha }{\cos \alpha }\right|}{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}\\ \Rightarrow p_3=\left|m^2\cos \alpha +2m\sin \alpha +\frac{{\sin }^2}{\cos \alpha }\right|\\ Hence,\\ \left(p_1,p_2,p_{3}areinG.P.\right)\end{array}$