If \((-4,\,3)\) and \((4,\,3)\) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
The third vertex is \(\bigl(0,\;3-4\sqrt{3}\bigr).\)
Step 1: Let the given vertices be \(A(-4,3)\) and \(B(4,3)\). Let the third vertex be \(C(x,y)\).
Step 2: In an equilateral triangle, all sides are equal, so:
\(AC = BC\) and also \(AB = AC\).
Step 3: First find \(AB\).
\(AB^2 = (4-(-4))^2 + (3-3)^2 = 8^2 + 0 = 64\)
So, \(AB = 8\). Hence \(AC = BC = 8\) and therefore \(AC^2 = BC^2 = 64\).
Step 4: Use \(AC = BC\) to get the relation between \(x\) and \(y\).
\(AC^2 = (x+4)^2 + (y-3)^2\) and \(BC^2 = (x-4)^2 + (y-3)^2\).
Since \(AC^2 = BC^2\):
\((x+4)^2 + (y-3)^2 = (x-4)^2 + (y-3)^2\)
Cancel \((y-3)^2\) from both sides:
\((x+4)^2 = (x-4)^2\)
Expand:
\(x^2 + 8x + 16 = x^2 - 8x + 16\)
\(16x = 0 \Rightarrow x = 0\).
Step 5: Now use \(AC = 8\) (i.e., \(AC^2 = 64\)) to find \(y\).
\(AC^2 = (x+4)^2 + (y-3)^2\)
Put \(x=0\):
\(64 = (0+4)^2 + (y-3)^2 = 16 + (y-3)^2\)
\((y-3)^2 = 48\)
\(y-3 = \pm \sqrt{48} = \pm 4\sqrt{3}\)
So \(y = 3 \pm 4\sqrt{3}\).
Step 6: Hence the two possible third vertices are:
\((0,\,3+4\sqrt{3})\) and \((0,\,3-4\sqrt{3})\).
Step 7: Use the given condition: the origin \((0,0)\) lies in the interior of the triangle.
If \(C=(0,3+4\sqrt{3})\), then all vertices have y-coordinate \(\ge 3\), so the triangle lies completely above the x-axis and cannot contain the origin.
Therefore we choose \(C=(0,3-4\sqrt{3})\), which makes the triangle extend downward and include \((0,0)\) inside.
Final Answer: The required third vertex is \(\boxed{(0,\,3-4\sqrt{3})}\).