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Find the line perpendicular to 3x+6y=12 that goes through (6,4)

$\begin{array}{rcl} 3x+6y &=& 12\\ 6y &=& 12-3x \\ y &=& \frac{12-3x}{6} \\ y &=& \frac{12}{6} -\frac{3x}{6} \\ y &=& 2 -\frac{x}{2} \\ y &=& -\frac{x}{2} + 2 \\ \end{array}\\$

$\begin{array}{rcl} \text{Formula } \boxed{~ \begin{array}{lrcl} y = mx+b \\\\ \dfrac{y-y_p}{x-x_p} = m_{\text{perpendicular}} \\ m_{\text{perpendicular}} = -\frac{1}{m} \end{array} ~}\\\\ \end{array}\\ \begin{array}{rcl} P(x_p,y_p) &=& (6,4) \\\\ y = -\frac{x}{2} + 2 \qquad m = -\frac12 \\ m_{\text{perpendicular}} = -\frac{1}{ -\frac12} &=& 2 \\ \dfrac{y-y_p}{x-x_p} = \dfrac{y-4}{x-6} &=& 2 \\ y-4 &=& 2\cdot (x-6) \\ y-4 &=& 2 x -12 \\ y &=& 2 x -12 +4 \\ y &=& 2 x - 8 \\ \end{array}\\$