waigaK

2021-08-13

Find all of the points on the x-axis which are 4 units from the point (-1, 1).

Corben Pittman

$\text{Let}\left\{\left({\left\{x\right\}}_{\left\{1\right\}},{\left\{y\right\}}_{\left\{1\right\}}\right)\right\}=\left\{\left(\left\{x\right\},\left\{0\right\}\right)\right\}\text{be the point on the x-axis. We also let}\left\{\left({\left\{x\right\}}_{\left\{2\right\}},{\left\{y\right\}}_{\left\{2\right\}}\right)\right\}=\left\{\left(-\left\{1\right\},\left\{1\right\}\right)\right\}.\text{Use the distance formula to find x where d=4:}$
$\left\{d\right\}=\sqrt{\left\{{\left\{\left({\left\{x\right\}}_{\left\{2\right\}}-{\left\{x\right\}}_{\left\{1\right\}}\right)\right\}}^{2}+{\left\{\left({\left\{y\right\}}_{\left\{2\right\}}-{\left\{y\right\}}_{\left\{1\right\}}\right)\right\}}^{2}\right\}}$
$\left\{4\right\}=\sqrt{\left\{{\left\{\left(-\left\{1\right\}-\left\{x\right\}\right)\right\}}^{2}+{\left\{\left(\left\{1\right\}-\left\{0\right\}\right)\right\}}^{2}\right\}}$
$\left\{4\right\}=\sqrt{\left\{\left\{1\right\}+\left\{2\right\}\left\{x\right\}+\left\{x\right\}\right\}}+{\left\{1\right\}}^{2}$
$\left\{4\right\}={\sqrt{\left\{y\right\}}}^{2}+\left\{6\right\}\left\{y\right\}+\left\{45\right\}$
$\text{Square both sides:}$
$\left\{16\right\}={\left\{x\right\}}^{2}+\left\{2\right\}\left\{x\right\}+\left\{2\right\}$
$\text{Subtract 36 from both sides:}$
$\left\{0\right\}={\left\{x\right\}}^{2}+\left\{2\right\}\left\{x\right\}-\left\{14\right\}$
$\text{Solve for x using quadratic formula:}$
$\left\{x\right\}=\left\{\frac{\left\{-\left\{b\right\}±\sqrt{\left\{{\left\{b\right\}}^{2}-\left\{4\right\}\left\{a\right\}\left\{c\right\}\right\}}\right\}}{\left\{\left\{2\right\}\left\{a\right\}\right\}}\right\}$
$\text{where a=1,b=2, and c=-14}$

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