Find the equation of the line through the point \( (10,-8) \) that is parallel to the line with equation \( y=-\frac{11}{6} x+21 \). The equation is

To find the equation of the line that is parallel to the given line, we first need to note that parallel lines share the same slope. The slope of the line \( y = -\frac{11}{6}x + 21 \) is \( -\frac{11}{6} \). Since we want the line passing through the point \( (10, -8) \) to have the same slope, we use point-slope form: \[ y - y_1 = m(x - x_1) \] Substituting \( m = -\frac{11}{6} \), \( x_1 = 10 \), and \( y_1 = -8 \): \[ y - (-8) = -\frac{11}{6}(x - 10) \] This simplifies to: \[ y + 8 = -\frac{11}{6}x + \frac{110}{6} \] So, \[ y = -\frac{11}{6}x + \frac{110}{6} - 8 \] Converting \( -8 \) to sixths gives us \( -48/6 \), resulting in: \[ y = -\frac{11}{6}x + \frac{110 - 48}{6} \] Thus, the final equation of the line is: \[ y = -\frac{11}{6}x + \frac{62}{6} \] Or equivalently: \[ y = -\frac{11}{6}x + \frac{31}{3} \]