Question

If two vertices of an equilateral triangle are (0,0) and (3, 0), find the third vertex​

Answer 1

$\huge\rm\underline\blue{GIVEN:}$

• $\large\rm{Two\:vertices\:of\:equilateral\:Triangle\:are}\:\\ \rm{(0,0)\:and\:(3,0)}$

$\huge\rm\underline\blue{TO\:FIND:}$

• $\large\rm{Third\:vertex\\:of\:equilateral\:Triangle}}$

$\huge\rm\underline\blue{SOLUTION:}$

$\large\rm{Let\:the\:third\:vertex\:C\:be\:(x_3,y_3)}$

$\large\rm{In\:Equilateral\:Triangle\:Length\:of\:all\:sides\:is\:equal}$

$\large\rm{Distance\:between\:those\:vertices\:should\:be\:equal}$

$\bold{\boxed{D\:=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}$

$\large\rm{D\rightarrow{Distance\:between\:any\:two\:points}}$

$\large\rm{Distance\:between\:(0,0)\:and\:(3,0)\:=\:\sqrt{(3-0)^2+(0-0)^2}}$

$\large\rm{\implies{D\:=\:\sqrt{9}}[\tex]</p><p></p><p>[tex]\large\rm{\implies{D\:=\:3\:units}}$

$\large\rm{So\:distance\:between\:AC\:=BC}$

$\large\rm{\sqrt{(x_2-x_3)^2+(y_2-y_3)^2\:=\:\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}}}$

$\large\rm{\implies{\sqrt{(x_3-3)^2+(y_3-0)^2}\:=\:\sqrt{(x_3-0)^2+(y_3-0)^2}}}$

$\large\rm{\implies{(x_3-3)^2+(y_3)^2\:=\:(x_3)^2+(y_3)^2}}$

$\bold{\boxed{(a-b)^2\:=\:a^2+b^2-2ab}}$

$\large\rm{\implies{(x_3)^2+9-2(x_3)(3)\:=\:(x_3)^2}}$

$\large\rm{\implies{9-6x_3\:=\:0}}$

$\large\rm{\implies{-6x_3\:=\:-9}}$

$\large\rm{\implies{x_3\:=\:-\frac{3}{2}}}$

$\large\rm{Distance\:between\:A\:and\:C\:=\:\sqrt{(x_3-x_1)^2+(y_3-y_1)^2}\:=\:9}}$

$\large\rm{\implies{\sqrt{(\frac{3}{2}-0)^2+(y_3-0)^2\:=\:3}}}$

$\large\rm{\implies{\sqrt{(\frac{9}{4})+(y_3)^2)\:=\:3}}}$

$\large\rm{Squaring\:on\:both\:sides}$

$\large\rm{\implies{\frac{9}{4}+(y_3)^2\:=\:9}}$

$\large\rm{\implies{(y_3)^2\:=\:9-\frac{9}{4}}}$

$\large\rm{\implies{(y_3)^2\:=\:\frac{25}{9}}}$

$\large\rm{\implies{y_3\:=\:\frac{5}{3}}}$

$\large\rm{\therefore{(x_3,y_3)\:=\:(\frac{3}{2},\frac{5}{3})}}$