If two vertices of a equilateral triangle are (3,0);(6,0). Find third

1.	If two vertices of a equilateral triangle are (3,0);(6,0). Find third vertex.
Answer» 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=ACBy distance formula we get,AB=BC=AC=3unitsAC=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 =\xa0{tex}(9 / 2,3 \\sqrt{\\frac{3}{2}}){/tex}

Discussion