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 $n^{t h}$ foot of the perpendicular on OA then $O P_{n}$ is equal to

{(\frac{1}{2})}^{n - 1}

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{(\frac{1}{2})}^{n}

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{(\frac{1}{2})}^{n + 1}

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None of these

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Solution

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

$x + y = 1 A (1, 0) B (0, 1) P (\frac{1}{2}, \frac{1}{2}) P_{1} (\frac{1}{2}, 0) M_{1} (\frac{1}{4}, \frac{1}{4}) P_{2} (\frac{1}{4}, 0) M_{2} (\frac{1}{8}, \frac{1}{8}) . . . . . P_{3} = (\frac{1}{8}, 0) P_{n} = [\frac{1}{2^{n}}, 0] M_{n} = [\frac{1}{2^{n + 1}}, \frac{1}{2^{n + 1}}] O P = \frac{1}{2^{n}} = {(\frac{1}{2})}^{n}$

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Similar questions

Q. 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^{t h}

foot of the perpendicular on OA from

M_{n - 1},

then O

P_{n}

is equal to

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 $O P_{n}$ =