# Using analytic geometry, find the coordinates of the point on

Using analytic geometry, find the coordinates of the point on the x axis which is equaidistant from and
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trisanualb6

Step 1
Given,

Let, Equidistent point,
So, using distance formula:
$\sqrt{{\left(x-3\right)}^{2}+{\left(-8\right)}^{2}}$=$\sqrt{\left(x+3{\right)}^{2}+\left(-4{\right)}^{2}}$
Both side
$\text{⧸}{x}^{2}+25-10x+64=\text{⧸}{x}^{2}+9+6x+16$
$25+64-9-16=10x+6x$
$16x=64$
$x=4$

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twineg4

Step 1
To find the distance between these two points we will use the distance formula
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
${d}_{1}=\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}$ and ${d}_{2}=\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}$
We have that ${d}_{1}={d}_{2}$ so we have:
$\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}$
Square both sides
${\left(\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}\right)}^{2}={\left(\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}\right)}^{2}$
Transform $\sqrt{{\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}}$ and $\sqrt{{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}}$ with using rule:
$\sqrt{a}={a}^{\frac{1}{2}}$
$\left(\left(\left(-5-x{\right)}^{2}+\left(2-0{\right)}^{2}{\right)}^{\frac{1}{2}}{\right)}^{2}=\left(\left(\left(2-x{\right)}^{2}+\left(3-0{\right)}^{2}{\right)}^{\frac{1}{2}}{\right)}^{2}$
Apply exponenet rule: ${\left({a}^{b}\right)}^{c}={a}^{bc}$
$\left({\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}\right)6\left\{\frac{1}{2}×2\right\}={\left({\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}\right)}^{\frac{1}{2}×2}$
Cancel th ecommon term into the exponent.
${\left(-5-x\right)}^{2}+{\left(2-0\right)}^{2}+{\left(2-x\right)}^{2}+{\left(3-0\right)}^{2}$
Subtract 2-0 and 2-0
${\left(-5-x\right)}^{2}+{2}^{2}={\left(2-x\right)}^{2}+{3}^{2}$
Step 2
Calculate exponents