16 The points \( A \) and \( B \) have the coordinates \( (4,6) \) and \( (12,2) \) respectively. The straight line \( l_{1} \) passes through \( A \) and \( B \). (a) Find an equation for \( l_{1} \) in the form \( a x+b y=c \), where \( a, b \) and \( c \) are integers.

To find the equation of the line \( l_1 \) that passes through the points \( A(4, 6) \) and \( B(12, 2) \), we first need to determine the slope of the line. The slope \( m \) of the line through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting the coordinates of points \( A \) and \( B \): \[ m = \frac{2 - 6}{12 - 4} = \frac{-4}{8} = -\frac{1}{2} \] Now that we have the slope, we can use the point-slope form of the equation of a line, which is: \[ y - y_1 = m(x - x_1) \] Using point \( A(4, 6) \): \[ y - 6 = -\frac{1}{2}(x - 4) \] Now, we simplify this equation: \[ y - 6 = -\frac{1}{2}x + 2 \] Adding 6 to both sides: \[ y = -\frac{1}{2}x + 8 \] To convert this into the form \( ax + by = c \), we can multiply through by 2 to eliminate the fraction: \[ 2y = -x + 16 \] Rearranging gives: \[ x + 2y = 16 \] Thus, the equation of the line \( l_1 \) in the form \( ax + by = c \) is: \[ \boxed{x + 2y = 16} \]