Question

\displaystyle\text{Find the point on the y-axis that is 6 units

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asked 2021-08-15
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\text{Find the point on the y-axis that is 6 units from the point }\ \ {\left\lbrace{\left(-{\left\lbrace{6}\right\rbrace},-{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}\)

Expert Answers (1)

2021-08-16
\(\displaystyle{d}{i}{s}{p}{l}{a}{y}{s}{t}{y}\le\text{Let}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}_{{{\left\lbrace{1}\right\rbrace}}},{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{1}\right\rbrace}}}\right)}\right\rbrace}={\left\lbrace{\left({\left\lbrace{0}\right\rbrace},{\left\lbrace{y}\right\rbrace}\right)}\right\rbrace}\text{be the point on the y-axis. We also let}{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}_{{{\left\lbrace{2}\right\rbrace}}},{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{2}\right\rbrace}}}\right)}\right\rbrace}={\left\lbrace{\left(-{\left\lbrace{6}\right\rbrace},-{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}.\)
\(\displaystyle\text{Use the distance formula to find y where d=6:}\)
\(\displaystyle{\left\lbrace{d}\right\rbrace}=\sqrt{{{\left\lbrace{\left\lbrace{\left({\left\lbrace{x}\right\rbrace}_{{{\left\lbrace{2}\right\rbrace}}}-{\left\lbrace{x}\right\rbrace}_{{{\left\lbrace{1}\right\rbrace}}}\right)}\right\rbrace}^{{{2}}}+{\left\lbrace{\left({\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{2}\right\rbrace}}}-{\left\lbrace{y}\right\rbrace}_{{{\left\lbrace{1}\right\rbrace}}}\right)}\right\rbrace}^{{{2}}}\right\rbrace}}}\)
\(\displaystyle{\left\lbrace{6}\right\rbrace}=\sqrt{{{\left\lbrace{\left\lbrace{\left(-{\left\lbrace{6}\right\rbrace}-{\left\lbrace{0}\right\rbrace}\right)}\right\rbrace}^{{{2}}}+{\left\lbrace{\left(-{\left\lbrace{3}\right\rbrace}-{\left\lbrace{y}\right\rbrace}\right)}\right\rbrace}^{{{2}}}\right\rbrace}}}\)
\(\displaystyle{\left\lbrace{6}\right\rbrace}=\sqrt{{{\left\lbrace{36}\right\rbrace}}}+{\left\lbrace{9}\right\rbrace}+{\left\lbrace{6}\right\rbrace}{\left\lbrace{y}\right\rbrace}+{\left\lbrace{y}\right\rbrace}^{{{2}}}\)
\(\displaystyle{\left\lbrace{6}\right\rbrace}=\sqrt{{{\left\lbrace{y}\right\rbrace}}}^{{{2}}}+{\left\lbrace{6}\right\rbrace}{\left\lbrace{y}\right\rbrace}+{\left\lbrace{45}\right\rbrace}\)
\(\displaystyle\text{Square both sides:}\)
\(\displaystyle{\left\lbrace{36}\right\rbrace}={\left\lbrace{y}\right\rbrace}^{{{2}}}+{\left\lbrace{6}\right\rbrace}{\left\lbrace{y}\right\rbrace}+{\left\lbrace{4}\right\rbrace}\)
\(\displaystyle\text{Subtract 36 from both sides}:\)
\(\displaystyle{\left\lbrace{0}\right\rbrace}={\left\lbrace{y}\right\rbrace}^{{{2}}}+{\left\lbrace{6}\right\rbrace}{\left\lbrace{y}\right\rbrace}+{\left\lbrace{9}\right\rbrace}\)
\(\displaystyle\text{Factor}:\)
\(\displaystyle{\left\lbrace{0}\right\rbrace}={\left\lbrace{\left({\left\lbrace{y}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}{\left\lbrace{\left({\left\lbrace{y}\right\rbrace}+{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}\)
\(\displaystyle\text{By zero product property}\),
\(\displaystyle{\left\lbrace{y}\right\rbrace}=-{\left\lbrace{3}\right\rbrace}\)
\(\displaystyle\text{So, the point that is 6 units from (-6,-3)is}:\)
\(\displaystyle{\left\lbrace{\left({\left\lbrace{0}\right\rbrace},-{\left\lbrace{3}\right\rbrace}\right)}\right\rbrace}\)
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