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The line \mathrm{x+y=1} meets \mathrm{x}-axis at \mathrm{A} and \mathrm{y}-axis at \mathrm{B . P} is the mid-point of \mathrm{A B} (fig.) \mathrm{P_1} is the foot of the perpendicular from \mathrm{P} to \mathrm{OA} ; \mathrm{M}_1 is that of \mathrm{P}_1 on \mathrm{OP; \mathrm{P}_2} is that of \mathrm{M}_1 on \mathrm{OA; \mathrm{M}_2} is that of \mathrm{\mathrm{P}_2} on O P ; \mathrm{P}_3 is that of \mathrm{M_2} from \mathrm{O A\ \&} so on. If \mathrm{P_n} denotes the \mathrm{n}th foot of the perpendicular on \mathrm{O A} from \mathrm{M_n-1}, then Find \mathrm{O P_n.}

Option: 1

\mathrm{\frac{1}{2^{n-1}}}


Option: 2

\mathrm{\frac{1}{3^{n}}}


Option: 3

\mathrm{\left ( \frac{2}{3} \right )^{n}}


Option: 4

\mathrm{\frac{1}{2^{n}}}


Answers (1)

best_answer

Let \mathrm{x+y=1} meets \mathrm{x}- axis at \mathrm{A\left ( 1,0 \right )} and \mathrm{y}- axis at \mathrm{B\left ( 0,1 \right )}
Also, \mathrm{\mathrm{OP}^2-1=O M_{n-1}^2+P_{n-1}+P_{n-1} M_{n-1}^2}
The coordinates of \mathrm{P} are \mathrm{\left ( 1/2,1/2 \right )} and \mathrm{PP_{1}} is perpendicular to \mathrm{OA.}
\mathrm{\Rightarrow \mathrm{OP}_1=\mathrm{P}_1 \mathrm{P}=1 / 2}                                                                                 
Equation of line \mathrm{\Rightarrow \mathrm{OP}} is \mathrm{y=x.}
We have \mathrm{O M_{n-1}^2=O P_n^2+P_n M_{n-1}^2=2 O_n^2=2 P_{n}^{2} (say)}                            
\mathrm{\begin{aligned} & =2 P_n^2+1 / 2 p_{n-1}^2 \quad \Rightarrow p^2=1 / 4 p_{n-1}^2 \Rightarrow p_n=1 / 2 p_{n-1} \\ & \therefore O_n=P_n=\frac{1}{2} p_{n-1}=\frac{1}{2^2} p_{n-2}=\ldots . .=\frac{1}{2^{n-1}} p_1=\frac{1}{2^n} \end{aligned}}

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rishi.raj

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