Standard Form of an Equation

**chrozer** · March 17th 2008, 02:02 PM

Write in standard form, the equation with integral coefficients for the set of coplanar points described. For each point, its distance from the fixed point $(-1,-2)$ is $\frac{1}{4}$ its distance from the fixed point $(6,-4)$ .

Is this answer right - $15x^2 + 15y^2 + 44x + 56y + 28 = 0$ ?

**galactus** · March 17th 2008, 02:23 PM

Let the point be (x,y).

We can use the distance formula.

$\sqrt{(x+1)^{2}+(y+2)^{2}}=\frac{1}{4}\sqrt{(x-6)^{2}+(y+4)^{2}}$

**chrozer** · March 17th 2008, 02:29 PM

Originally Posted by galactus

Let the point be (x,y).

We can use the distance formula.

$(x+1)^{2}+(y+2)^{2}=\frac{1}{4}\left[(x-6)^{2}+(y+4)^{2}\right]$

Should't the equation be $(x+1)^{2}+(y+2)^{2}=\frac{1}{16}\left[(x-6)^{2}+(y+4)^{2}\right]$ instead. After your square both sides to get rid of the radical you also square $\frac {1}{4}$ to get $\frac {1}{16}$ and from there on solve. So is my answer right in my 1st post?

**galactus** · March 17th 2008, 02:41 PM

Yes, you're correct. I forgot to include my radical signs.

**chrozer** · March 17th 2008, 03:00 PM

Originally Posted by galactus

Yes, you're correct. I forgot to include my radical signs.

Alright thnx. I just wanted to make sure I answered the equation right?

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