Two vertices of an equilateral triangle are (a,-a) and (-a,a). Find the third.

Friday, March 6, 2015

Two vertices of an equilateral triangle are (a,-a) and (-a,a). Find the third.

Hello!

An equilateral triangle, by definition, has all three sides of equal lengths.

For two points with the coordinates $\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}$ and $\displaystyle{\left({x}_{{2}},{y}_{{2}}\right)}$ the distance between them is

$\displaystyle{d}=\sqrt{{{{\left({x}_{{1}}-{x}_{{2}}\right)}}^{{2}}+{{\left({y}_{{1}}-{y}_{{2}}\right)}}^{{2}}}}.$

The length of the given side is $\displaystyle\sqrt{{{{\left({2}{a}\right)}}^{{2}}+{{\left(-{2}{a}\right)}}^{{2}}}}={2}{\left|{a}\right|}\sqrt{{{2}}}.$

Denote the third point coordinates as $\displaystyle{x}$ and $\displaystyle{y},$ then the lengths of the other side are

$\displaystyle{{d}_{{1}}^{{2}}}=\sqrt{{{{\left({x}-{a}\right)}}^{{2}}+{{\left({y}+{a}\right)}}^{{2}}}}$ and $\displaystyle{{d}_{{2}}^{{2}}}=\sqrt{{{{\left({x}+{a}\right)}}^{{2}}+{{\left({y}-{a}\right)}}^{{2}}}}.$

And $\displaystyle{d}_{{1}}={d}_{{2}}={2}{\left|{a}\right|}\sqrt{{{2}}}.$ This gives us the system of two equations with two variables.

Both sides of both equations are nonnegative, so we can square both equations and will obtain

$\displaystyle{8}{{a}}^{{2}}={{x}}^{{2}}-{2}{a}{x}+{{a}}^{{2}}+{{y}}^{{2}}+{2}{a}{y}+{{a}}^{{2}},$
$\displaystyle{8}{{a}}^{{2}}={{x}}^{{2}}+{2}{a}{x}+{{a}}^{{2}}+{{y}}^{{2}}-{2}{a}{y}+{{a}}^{{2}}.$

Subtract the first from the second and obtain

$\displaystyle{0}={4}{a}{x}-{4}{a}{y},$ or $\displaystyle{x}={y}$ (if $\displaystyle{a}\ne{0}$ ).

Now substitute this into the first equation and obtain

$\displaystyle{8}{{a}}^{{2}}={2}{{x}}^{{2}}+{2}{{a}}^{{2}},$ or $\displaystyle{{x}}^{{2}}={3}{{a}}^{{2}},$ or $\displaystyle{x}=\pm{\left|{a}\right|}\sqrt{{{3}}}.$

So for $\displaystyle{a}\ne{0}$ there are two solutions, $\displaystyle{\left({\left|{a}\right|}\sqrt{{{3}}},{\left|{a}\right|}\sqrt{{{3}}}\right)}$ and $\displaystyle{\left(-{\left|{a}\right|}\sqrt{{{3}}},-{\left|{a}\right|}\sqrt{{{3}}}\right)}.$

For $\displaystyle{a}={0}$ the only solution is $\displaystyle{\left({0},{0}\right)}.$

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Friday, March 6, 2015

Two vertices of an equilateral triangle are (a,-a) and (-a,a). Find the third.

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