Two vertices of an equilateral triangle are $(0,0)$ and $(3,\sqrt3)$. Find the coordinates of the third vertex (there are two possible answers; find the one with a positive $y$-coordinate).
Step-by-Step Solution
Key Concept: The third vertex must be equidistant from both given points, at the same distance as the side length; use the distance formula to set up two equations.
Side length $=\sqrt{9+3}=\sqrt{12}=2\sqrt3$. Let the third vertex be $(x,y)$: $x^2+y^2=12$ and $(x-3)^2+(y-\sqrt3)^2=12$. [1.5 Marks]
Expanding the second equation and subtracting the first: $-6x+9-2\sqrt3y+3=0\Rightarrow6x+2\sqrt3y=12\Rightarrow3x+\sqrt3y=6$. [1.0 Mark]
Solving this together with $x^2+y^2=12$ gives $(x,y)=(0,2\sqrt3)$ (taking the positive-$y$ solution; the other solution is $(3,-\sqrt3)$). [0.5 Mark]
Correct Answer: