If (-4,3) and (4,3) are two vertices of an eq | vedclass

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(D) Let the vertices of the equilateral triangle be $A(-4,3)$,$B(4,3)$,and $C(x,y)$.Since the triangle is equilateral,$AB = BC = CA$,which implies $AB

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$(0,3-4 \sqrt{3})$

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(D) Let the vertices of the equilateral triangle be $A(-4,3)$,$B(4,3)$,and $C(x,y)$.Since the triangle is equilateral,$AB = BC = CA$,which implies $AB^2 = BC^2 = CA^2$.First,calculate $AB^2 = (4 - (-4))^2 + (3 - 3)^2 = 8^2 + 0^2 = 64$.For $BC^2 = CA^2$,we have $(x-4)^2 + (y-3)^2 = (x+4)^2 + (y-3)^2$.This simplifies to $(x-4)^2 = (x+4)^2$,which gives $x^2 - 8x + 16 = x^2 + 8x + 16$,so $16x = 0$,hence $x = 0$.Now,use $AB^2 = BC^2$ with $x=0$: $64 = (0-4)^2 + (y-3)^2$.$64 = 16 + (y-3)^2$,so $(y-3)^2 = 48$.$y-3 = \pm \sqrt{48} = \pm 4\sqrt{3}$.Thus,$y = 3 \pm 4\sqrt{3}$.The two possible vertices are $C_1(0, 3+4\sqrt{3})$ and $C_2(0, 3-4\sqrt{3})$.The origin $(0,0)$ lies in the interior of the triangle. For the triangle with vertices $(-4,3)$,$(4,3)$,and $(0, y)$,the $y$-coordinate of the third vertex must be less than $3$ for the origin to be inside. Since $3+4\sqrt{3} > 3$ and $3-4\sqrt{3} < 3$,the correct vertex is $(0, 3-4\sqrt{3})$.

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.

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