18. A has coordinates \( (-3,0) \) \( B \) has coordinates \( (1,6) \) \( C \) has coordinates \( (5,2) \) Find an equation of the line that passes through \( C \) and is perpendicular to \( A B \). Give your equation in the form \( a x+b y=c \) where \( a, b \) and \( c \) are integers.

To find the equation of the line that passes through point \( C(5,2) \) and is perpendicular to line \( AB \), we first need to determine the slope of \( AB \). The slope \( m_{AB} \) between points \( A(-3,0) \) and \( B(1,6) \) is calculated as follows: \[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 0}{1 - (-3)} = \frac{6}{4} = \frac{3}{2}. \] The slope of the line perpendicular to \( AB \) is the negative reciprocal of \( m_{AB} \): \[ m_{\text{perpendicular}} = -\frac{1}{m_{AB}} = -\frac{2}{3}. \] Now we use the point-slope form of the equation of a line to find the equation of the line through point \( C(5,2) \): \[ y - y_1 = m(x - x_1). \] Substituting in \( C(5,2) \) and the slope \( -\frac{2}{3} \): \[ y - 2 = -\frac{2}{3}(x - 5). \] Now, simplify this equation: \[ y - 2 = -\frac{2}{3}x + \frac{10}{3}. \] Adding 2 (which is \( \frac{6}{3} \)) to both sides gives us: \[ y = -\frac{2}{3}x + \frac{10}{3} + \frac{6}{3} = -\frac{2}{3}x + \frac{16}{3}. \] To convert this to the form \( ax + by = c \), we can multiply through by 3 to eliminate the fractions: \[ 3y = -2x + 16. \] Rearranging gives: \[ 2x + 3y = 16. \] Thus, the equation of the line in the form \( ax + by = c \) is: \[ \boxed{2x + 3y = 16}. \]