Right triangle whose sides are 3cm …

Let ABC be the equilateral triangle such that,
A = (3,0), B=(6,0) and C=(x,y)
Distance between:
{tex}\sqrt {( x_{2}-x_{1})^2+(y_{2} -y_{1})^{2} }{/tex}
we know that,
AB=BC=AC
By distance formula we get,
AB=BC=AC=3units
AC=BC
{tex}\sqrt{(3-x)^2+y^2}=\sqrt{(6-x)^2+y^2}{/tex}
{tex}9+x^2-6 x+y^2=36+x^2-12 x+y^2{/tex}
{tex}6 x=27{/tex}
{tex}x=27 / 6=9 / 2{/tex}
BC = 3 units
{tex}\sqrt{(6-\frac{27}{6})^2+y^2}=3{/tex}
{tex}(\frac{(36-27)}{6})^2+y^2=9{/tex}
{tex}(\frac{9}{6})^2+y^2=9{/tex}
{tex}(\frac{3}{2})^2+y^2=9{/tex}
{tex}\frac{9}{4}+y^2=9{/tex}
{tex}9+4 y^2=36{/tex}
{tex}4 y^2=27{/tex}
{tex}y^2=\frac{27}{4}{/tex}
{tex}y=\sqrt{(\frac{27}{4})}{/tex}
{tex}y=3 \sqrt{\frac{3}{2}}{/tex}
{tex}(x, y)=(9 / 2,3 \sqrt{\frac{3}{2}}){/tex}
Hence third vertex of equilateral triangle = C = {tex}(9 / 2,3 \sqrt{\frac{3}{2}}){/tex}