Question

The line x + y = 1 meets x-axis at A and y axis at B, P is the mid point of AB, $P_1$ is the foot of the perpendicular from P to OA; $M_1$ is that of $P_1$ from OP; $P_2$ is that of $M_1$ from OA; $M_2$ is that of $P_2$ from OP; $P_3$ is that of $M_2$ from OA and so on. If $P_n$ denotes the $n^{th}$ foot of the perpendicular on OA from $M_{n-1},$ then O$P_n$ is equal to

• A. $\frac{1}{2n}$
• B. $\frac{1}{2^n}$
• C. $2^n-1$
• D. $2^n+1$

Answer 1

The correct option is B $\frac{1}{2^n}$

Equation of AB is x + y = 1, then coordinates of A and B are (1, 0) and (0, 1) respectively.
$\therefore$ coordinates of P is $(\frac{1}{2},\frac{1}{2})$
$P{P}_1$ is perpendicular to OA
Equation OP is y = x
$\Rightarrow O{P}_1=P{P}_1=\frac{1}{2}$

We have $(O{M}_{n-1}{)}^2=(O{P}_n{)}^2+(P_n{M}_{n-1}{)}^2$
$=2(O{P}_n{)}^2$ $\{\because y=x)$
$=2(O{P}_n{)}^2=2{\alpha }_n^2$ $\{whereweletO{P}_n={\alpha }_n\}$
We also have
$(O{P}_{n-1}{)}^2=(O{M}_{n-1}{)}^2+(P_{n-1}{M}_{n-1}{)}^2{\alpha }_{n-1}^2=2{\alpha }_n^2+\frac{1}{2}{\alpha }_{n-1}^2\frac{1}{2}{\alpha }_{n-1}=2{\alpha }_n^2{\alpha }_n=\frac{1}{2}{\alpha }_{n-1}=\frac{1}{2^2}{\alpha }_{n-2}=\frac{1}{2^3}{\alpha }_{n-3}=...=\frac{1}{2^{n-1}}\frac{1}{2}=\frac{1}{2^n}$