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
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
$\displaystyle \sqrt{(x+3)^2+(y+2)^2}=2*(\sqrt{(x+\frac{3}{2})^2+ (y-y)^2})$
$\displaystyle x^2+6x+9+y^2+4y+4=2x^2+6x+9/4+0$
$\displaystyle -x^2+y^2+0x+4y+17/2=0$