Tobias Ali

2021-08-15

Find the point on the y-axis that is 6 units from the point $\left\{\left(-\left\{6\right\},-\left\{3\right\}\right)\right\}$

### Answer & Explanation

Elberte

$\text{Let}\left\{\left({\left\{x\right\}}_{\left\{1\right\}},{\left\{y\right\}}_{\left\{1\right\}}\right)\right\}=\left\{\left(\left\{0\right\},\left\{y\right\}\right)\right\}\text{be the point on the y-axis. We also let}\left\{\left({\left\{x\right\}}_{\left\{2\right\}},{\left\{y\right\}}_{\left\{2\right\}}\right)\right\}=\left\{\left(-\left\{6\right\},-\left\{3\right\}\right)\right\}.$
$\text{Use the distance formula to find y where d=6:}$
$\left\{d\right\}=\sqrt{\left\{{\left\{\left({\left\{x\right\}}_{\left\{2\right\}}-{\left\{x\right\}}_{\left\{1\right\}}\right)\right\}}^{2}+{\left\{\left({\left\{y\right\}}_{\left\{2\right\}}-{\left\{y\right\}}_{\left\{1\right\}}\right)\right\}}^{2}\right\}}$
$\left\{6\right\}=\sqrt{\left\{{\left\{\left(-\left\{6\right\}-\left\{0\right\}\right)\right\}}^{2}+{\left\{\left(-\left\{3\right\}-\left\{y\right\}\right)\right\}}^{2}\right\}}$
$\left\{6\right\}=\sqrt{\left\{36\right\}}+\left\{9\right\}+\left\{6\right\}\left\{y\right\}+{\left\{y\right\}}^{2}$
$\left\{6\right\}={\sqrt{\left\{y\right\}}}^{2}+\left\{6\right\}\left\{y\right\}+\left\{45\right\}$
$\text{Square both sides:}$
$\left\{36\right\}={\left\{y\right\}}^{2}+\left\{6\right\}\left\{y\right\}+\left\{4\right\}$
$\text{Subtract 36 from both sides}:$
$\left\{0\right\}={\left\{y\right\}}^{2}+\left\{6\right\}\left\{y\right\}+\left\{9\right\}$
$\text{Factor}:$
$\left\{0\right\}=\left\{\left(\left\{y\right\}+\left\{3\right\}\right)\right\}\left\{\left(\left\{y\right\}+\left\{3\right\}\right)\right\}$
$\text{By zero product property}$,
$\left\{y\right\}=-\left\{3\right\}$
$\text{So, the point that is 6 units from (-6,-3)is}:$
$\left\{\left(\left\{0\right\},-\left\{3\right\}\right)\right\}$

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