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; is that from $P_2$ to OP; $P_3$ is that from $M_1$ 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.
• B.
• C.
• D.

Answer 1

The correct option is B

x + y = 1 meets x-axis at A(1, 0) and y-axis at B(0, 1).

The coordinated of P are $(\frac{1}{2},\frac{1}{2})$ and $P{P}_1$ is perpendicular to OA.
$\Rightarrow O{P}_1=P_{1}P=1/2$
Equation of line OP is y = x
We have $(O{M}_{n-1}{)}^2=(O{P}_n{)}^2+(P_n{M}_{n-1}{)}^2=2(O{P}_n{)}^2=2P_n^2\left(say\right)$
Also, $(O{P}_{n-1}{)}^2=(O{M}_{n-1}{)}^2+(P_{n-1}{M}_{n-1}{)}^2$
$(O{P}_{n-1}{)}^2=(O{M}_{n-1}{)}^2+(P_{n-1}{M}_{n-1}{)}^2=2p_n^2+\frac{1}{2}{p}_{n-1}^2$
$\Rightarrow p_n^2-{\frac{1}{4}}_{n-1}^2\Rightarrow p_n=\frac{1}{2}{p}_{n-1}$​
$\therefore O{P}_n=p_n=\frac{1}{2}{p}_{n-1}=\frac{1}{2^2}{p}_{n-2}=....=\frac{1}{2^{n-1}}{p}_1=\frac{1}{2^n}$