1. ## Loci

Determine the locus of the point P(x,y) which moves so that its distance from (-3,-2) is twice its perpendicular distance from 2x+3=0

2. ## Re: Loci

Are you sure it's 2x + 3 = 0 and not 2x + 3 = y?

3. ## Re: Loci

$\displaystyle \sqrt{(x+3)^2+(y+2)^2}=2\left|\dfrac{2x+3}{\sqrt{2 ^2+0^2}}\right|$ now, take squares.

Edited: Sorry, I didn't see Prove It's post.

4. ## Re: Loci

You have to use the same formula on each side this is how i did mines sorry i was unable to load it because it was thundering outside

I first solve for x so x=-3/2 and y is just y so we have the point (-3/2,y)

Using the distance formula

$\displaystyle \sqrt{(x+3)^2+(y+2)^2}=2*(\sqrt{(x+\frac{3}{2})^2+ (y-y)^2})$

$\displaystyle x^2+6x+9+y^2+4y+4=2x^2+6x+9/4+0$

$\displaystyle -x^2+y^2+0x+4y+17/2=0$

5. ## Re: Loci

If it is, in fact, the vertical line 3x+2= 0 or x= -2/3, then we have
$\displaystyle \sqrt{(x+3)^2+ (y+2)^2}= 2|x+ 3/2|$

Square both sides, combine like terms, and complete the square. I believe this is a hyperbola.