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(C) Let the slope of the required lines be $m$. The given line is $x + y = 2$,which has a slope of $m_1 = -1$.Since the triangle is equilateral,the an

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$y - 3 = (2 \pm \sqrt{3})(x - 2)$

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(C) Let the slope of the required lines be $m$. The given line is $x + y = 2$,which has a slope of $m_1 = -1$.Since the triangle is equilateral,the angle between the sides is $60^\circ$.Using the formula $	an 	heta = |\frac{m - m_1}{1 + m \cdot m_1}|$,we have:$	an 60^\circ = |\frac{m - (-1)}{1 + m(-1)}|$$\sqrt{3} = |\frac{m + 1}{1 - m}|$This gives two cases:$1) \sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow \sqrt{3} - \sqrt{3}m = m + 1$ $\Rightarrow m(1 + \sqrt{3}) = \sqrt{3} - 1$ $\Rightarrow m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3}$$2) -\sqrt{3} = \frac{m + 1}{1 - m}$ $\Rightarrow -\sqrt{3} + \sqrt{3}m = m + 1$ $\Rightarrow m(\sqrt{3} - 1) = \sqrt{3} + 1$ $\Rightarrow m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 3)$:$y - 3 = (2 \pm \sqrt{3})(x - 2)$.

One vertex of an equilateral triangle is $(2, 3)$ and the line of its opposite side is $x + y = 2$. Find the equations of the other two sides.

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