A-point-P-x-y-moves-such-that-its-perpendicular-distance-from-the-line-12x-5y-1-0-is-always-3-units-Find-the-equation-that-describes-the-locus-precisely-

Question Number 150738 by nadovic last updated on 15/Aug/21

A point P (x, y) moves such that its

perpendicular distance from the line

12 x + 5 y - 1 = 0 is always 3 u n i t s .

Find the equation that describes the

locus precisely .

Commented by nadovic last updated on 15/Aug/21

T h a n k s S i r

Commented by JDamian last updated on 15/Aug/21

In other words, find the (two) parallel lines whose distances from the given line are 3 u.

Answered by nimnim last updated on 15/Aug/21

y = - \frac{12}{5} x + \frac{1}{5} \Rightarrow m = - \frac{12}{5}, C_{1} = \frac{1}{5}

L e t t h e e q u a t i o n b e y = m x + C_{2} \dots \dots (i)

∴ 3 = \frac{∣ C_{1} - C_{2} ∣}{\sqrt{1 + m^{2}}} \Rightarrow 3 = \frac{∣ \frac{1}{5} - C_{2} ∣}{13 / 5}

\Rightarrow \frac{39}{5} =∣ \frac{1}{5} - C_{2} ∣

\Rightarrow C_{2} = \frac{1}{5} - \frac{39}{5} o r C_{2} = \frac{39}{5} + \frac{1}{5}

\Rightarrow C_{2} = - \frac{38}{5} o r 8

f r o m (i) y = - \frac{12}{5} x - \frac{34}{5} a n d y = - \frac{12}{5} x + 8

\Rightarrow 12 x + 5 y + 34 = 0 a n d 12 x + 5 y - 40 = 0

Commented by nadovic last updated on 15/Aug/21

T h a n k y o u S i r

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