2 years ago

Can anyone help me with this question please . I need to turn this in a few minutes.

Mathematics

1 answer:

n200080 [17]2 years ago

3 0

Answer:

rip

Step-by-step explanation:

its been 5 min

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Step-by-step explanation:

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3 years ago

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.

leonid [27]

Answer-

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

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

Solution-

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

The applying the distance formula,

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

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

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

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

$\Rightarrow 12h-5k=\pm 14$

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

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

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

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