# Thread: How Would I Express This Equation Using the Distance Formula?

1. ## How Would I Express This Equation Using the Distance Formula?

For each point, its distance from the fixed point (-3,0) is twice its distance from the fixed point (3,0).

My book gives me the simplified answer, but I can't figure out how to express it in the distance formula to show my work.

Also, if someone's kind enough, please express the following as well: "For each point, its distance from the fixed point (4,3) is 3 times its distance from the fixed point (-1,2)."

FYI, this isn't my homework. I just need to know how to do this for an upcoming test.

Oh, and an explanation of where to plug in numbers would be greatly appreciated.

2. Originally Posted by Noraowolf1
For each point, its distance from the fixed point (-3,0) is twice its distance from the fixed point (3,0).

My book gives me the simplified answer, but I can't figure out how to express it in the distance formula to show my work.

Also, if someone's kind enough, please express the following as well: "For each point, its distance from the fixed point (4,3) is 3 times its distance from the fixed point (-1,2)."

FYI, this isn't my homework. I just need to know how to do this for an upcoming test.

Oh, and an explanation of where to plug in numbers would be greatly appreciated.
I can see it now...

Syllybus for Pre-Algebra:
1) How to do stuff for tests.
2) Where to plug stuff in.

That is a fine course.

Distance from (-3,0) is $\sqrt{(x-(-3))^{2}+(y-0)^{2}}$
Simplify a bit $\sqrt{(x+3)^{2}+y^{2}}$

Distance from (3,0) is $\sqrt{(x-3)^{2}+(y-0)^{2}}$
Simplify a bit $\sqrt{(x-3)^{2}+y^{2}}$

Twice the Distance
$\sqrt{(x+3)^{2}+y^{2}} = 2\cdot \sqrt{(x-3)^{2}+y^{2}}$

Everything is positive, so there is no harm in squaring both sides.
$(x+3)^{2}+y^{2} = 4\cdot ((x-3)^{2}+y^{2})$

Now what?

3. Originally Posted by TKHunny
I can see it now...

Syllybus for Pre-Algebra:
1) How to do stuff for tests.
2) Where to plug stuff in.

That is a fine course.

Distance from (-3,0) is $\sqrt{(x-(-3))^{2}+(y-0)^{2}}$
Simplify a bit $\sqrt{(x+3)^{2}+y^{2}}$

Distance from (3,0) is $\sqrt{(x-3)^{2}+(y-0)^{2}}$
Simplify a bit $\sqrt{(x-3)^{2}+y^{2}}$

Twice the Distance
$\sqrt{(x+3)^{2}+y^{2}} = 2\cdot \sqrt{(x-3)^{2}+y^{2}}$

Everything is positive, so there is no harm in squaring both sides.
$(x+3)^{2}+y^{2} = 4\cdot ((x-3)^{2}+y^{2})$

Now what?
Well, my book says the simplified answer is $x^2+y^2-10x+9=0$

Oh, and it's actually Advanced Algebra II. Not Pre-Alg.

4. Sweet! Take my final form and use your best algebra skills to get it to the book's form.

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