Using analytic geometry, find the coordinates of the point on

Step 1
To find the distance between these two points we will use the distance formula
${\displaystyle d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}}$
${\displaystyle d_1=\sqrt{{(-5-x)}^2+{(2-0)}^2}}$ and ${\displaystyle d_2=\sqrt{{(2-x)}^2+{(3-0)}^2}}$
We have that ${\displaystyle d_1=d_2}$ so we have:
${\displaystyle \sqrt{{(-5-x)}^2+{(2-0)}^2}=\sqrt{{(2-x)}^2+{(3-0)}^2}}$
Square both sides
${\displaystyle {\left(\sqrt{{(-5-x)}^2+{(2-0)}^2}\right)}^2={\left(\sqrt{{(2-x)}^2+{(3-0)}^2}\right)}^2}$
Transform ${\displaystyle \sqrt{{(-5-x)}^2+{(2-0)}^2}}$ and ${\displaystyle \sqrt{{(2-x)}^2+{(3-0)}^2}}$ with using rule:
${\displaystyle \sqrt{a}=a^{\frac{1}{2}}}$
$(((-5-x{)}^2+(2-0{)}^2{)}^{\frac{1}{2}}{)}^2=(((2-x{)}^2+(3-0{)}^2{)}^{\frac{1}{2}}{)}^2$
Apply exponenet rule: ${\displaystyle {\left(a^b\right)}^c=a^{bc}}$
${\displaystyle ({(-5-x)}^2+{(2-0)}^2)6\{\frac{1}{2}\times 2\}={({(2-x)}^2+{(3-0)}^2)}^{\frac{1}{2}\times 2}}$
Cancel th ecommon term into the exponent.
${\displaystyle {(-5-x)}^2+{(2-0)}^2+{(2-x)}^2+{(3-0)}^2}$
Subtract 2-0 and 2-0
${\displaystyle {(-5-x)}^2+2^2={(2-x)}^2+3^2}$
Step 2
Calculate exponents

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