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Rudik [331]
2 years ago
11

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

Mathematics
1 answer:
leonid [27]2 years ago
4 0

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

12x-5y+14=0 \\ 12x-5y-14=0

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1

\Rightarrow 12h-5k-1=\pm 13

\Rightarrow 12h-5k=\pm 14

\Rightarrow 12h-5k=14,\ 12h-5k=-14

\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0

\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0

Two equations are formed because one will be upper from the the given line and other will be below it.

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So, the expression becomes

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