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The equation of the line passing through the point (1, 0) and at a distance $32$ from the origin is:

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3 x + y − 3 = 0, 3 x − y − 3 = 0

3 x + y + 3 = 0, 3 x − y + 3 = 0

x + 3 y − 3 = 0, x − 3 y − 3 = 0

None of these

answer is A.

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Detailed Solution

We know that the equation of a line passing through the point

(x 1, y 1)

is given by

⇒ y − y 1 = m (x − x 1)

, where m is the slope of the line.
We have the point (1, 0) and if we compare it to

(x 1, y 1)

, then we get

x 1 = 1

and

y 1 = 0

So, putting these values in the above equation, we will get:-

⇒ y − 0 = m (x − 1)

Simplifying the equation to get the following equation:-

⇒ y − m x + m = 0 ⇒ m x + y − m = 0

We know that distance of the line

a x + b y + c = 0

from the origin is given by:

d = | c | a 2 + b 2

. Therefore, as we compare the both of equation, we have

a = m, b = 1 and c = − m

.
So, the distance from origin will be:

d = | − m | m 2 + 12

Since, we are already given that the distance of line from origin is

32 u n i t s .

∴ d = | − m | m 2 + 12 = 32 ⇒ | − m | m 2 + 12 = 32

Squaring both sides, we will get:-

⇒ | − m | m 2 + 12 2 = 322

⇒ m 2 m 2 + 1 = 34

Cross - multiplying the terms to get the following expression:-

⇒ 4 m 2 = 3 m 2 + 1

Simplifying the RHS by opening up the parenthesis on the RHS, we will then obtain:-

⇒ 4 m 2 = 3 m 2 + 3

⇒ 4 m 2 − 3 m 2 = 3

⇒ m 2 = 3 ⇒ m = ± 3

Putting this in (1), we will get-

⇒ 3 x + y − 3 = 0 or − 3 x + y + 3 = 0 ⇒ 3 x + y − 3 = 0 or 3 x − y − 3 = 0

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