How Would I Express This Equation Using the Distance Formula?

**Noraowolf1** · April 19th 2010, 06:58 PM

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.

**TKHunny** · April 19th 2010, 07:27 PM

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?

**Noraowolf1** · April 19th 2010, 07:48 PM

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.

**TKHunny** · April 19th 2010, 07:56 PM

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

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