## Loser66 one year ago $$\sqrt{(x+3)^2+(y-2)^2}=\sqrt{(x-3)^2+y^2}$$ In the xy plane, the set of points whose coordinates satisfy the equation above is?? 1) A line 2) A circle 3) An ellipse 4) A parabola 5) One branch of a hyperbola Please, help

1. ganeshie8

Notice the distance formulas on left and right hand sides, the given equation is same as :  distance between (x,y) and (-3, 2) = distance between (x, y) and (3, 0) 

2. ganeshie8

so you need to find the locus of points that stay at same distance from two points : (-3, 2) and (3, 0)

3. ganeshie8

Thats exactly the definition of perpendicular bisector!

4. Loser66

Got you, thank you so much.

5. Loser66

Damn!! They are distance formulas!! how can I not see it!!

6. ganeshie8

Haha solving them is fun too, square both sides and all quadratic terms cancel out.. gives the equation of perpendicular bisector..

7. Loser66

I did that way but stop at square both sides and look at the square of x and y, I think there must be other way to find the answer out quickly. So that I stopped and post the question. hehehe... I got it from you. Thank you so much

8. ganeshie8

np