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 from $P_1$ to OP; $P_2$ is that from $M_1$ to OA; $M_2$ is that from $P_2$ to OP; $P_3$ is that from $M_2$ to OA and so on. If $P_n$ denotes the nth foot of the perpendicular on OA from $M_{n-1}$, then $O{P}_n$ =

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

Answer 1

The correct option is B

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

$x+y=1$ meets x - axis at $A(1,0)$ and y - axis at $B(0,1)$
The coordinates of P are $(\frac{1}{2},\frac{1}{2})$ and $P{P}_1$ is perpendicular to OA.
$\Rightarrow O{P}_1=P_{1}P=\frac{1}{2}$
Equation of line OP is $y=x$.
We have $(O{M}_{n-1}{)}^2=(O{P}_n{)}^2+(P_n{M}_{n-1}{)}^2=2(O{P}_n{)}^2=2p_n^2$ (say)
Also, $(O{P}_{n-1}{)}^2=(O{M}_{n-1}{)}^2+(P_{n-1}{M}_{n-1}{)}^2$
$=2p_n^2+\frac{1}{2}{p}_{n-1}^2\Rightarrow p_n^2=\frac{1}{4}{p}_{n-1}^2\Rightarrow p_n=\frac{1}{2}{p}_{n-1}\therefore O{P}_n=p_n=\frac{1}{2}{p}_{n-1}=\frac{1}{2^2}{p}_{n-2}=\cdots =\frac{1}{2^{n-1}}{p}_1=\frac{1}{2^n}$