\(2x+4y=6\\ y = -\dfrac 1 2 x + \dfrac 3 2\\ m=-\dfrac 1 2 \\ m_\perp=-\dfrac{1}{m} = 2\\ \text{so using point slope form the equation defining the perpendicular line is}\\ (y-4)=2(x-6) \\ -2x+y = -8\)
\(2x+4y=6\\ y = -\dfrac 1 2 x + \dfrac 3 2\\ m=-\dfrac 1 2 \\ m_\perp=-\dfrac{1}{m} = 2\\ \text{so using point slope form the equation defining the perpendicular line is}\\ (y-4)=2(x-6) \\ -2x+y = -8\)
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