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

Friday, August 9, 2013

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...

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

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 $\displaystyle{\left(-{\left|{a}\right|}\sqrt{{{3}}},-{\left|{a}\right|}\sqrt{{{3}}}\right)}.$