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 and so on. If  $P_n$ denotes the $n^{th}$  foot of the perpendicular on OA from $M_{n-1}$ ¸ then $\frac{O{P}_{10}}{O{P}_8}$ =

(b) : $x+y=1$ 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
 ${\left(O{M}_{n-1}\right)}^2={\left(O{P}_n\right)}^2+{\left(P_n{M}_{n-1}\right)}^2=2{\left(O{P}_n\right)}^2=2p_n^2\left(say\right)$
Also,  ${\left(O{P}_{n-1}\right)}^2={\left(O{M}_{n-1}\right)}^2+{\left(P_{n-1}{M}_{n-1}\right)}^2$
$\Rightarrow {\left(O{P}_{n-1}\right)}^2=2p_n^2+\frac{1}{2}{p}_{n-1}^2$ 
 $\Rightarrow p_n^2=\frac{1}{4}{p}_{n-1}^2\Rightarrow p_n=\frac{1}{2}{p}_{n-1}$
 $\therefore O{P}_n=p_n=\frac{1}{2}{p}_{n-1}=\frac{1}{2^2}{p}_{n-2}=\mathrm{...}=\frac{1}{2^{n-1}}{p}_1=\frac{1}{2^n}$