Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through \( (1,6) \) and perpendicular to \( 3 x+7 y=1 \). a) The equation of the line in slope-intercept form is (Type your answer in slope-intercept form. Use integers or fractions for any numbers in the equation.) b) The equation of the line in standard form is (Type your answer in standard form.)

To find the equation of the line passing through the point \( (1,6) \) and perpendicular to the given line, we first need to determine the slope of the given line. The equation of the line is \( 3x + 7y = 1 \). We can rearrange this into slope-intercept form ( \( y = mx + b \) ): 1. Start by isolating \( y \): \[ 7y = -3x + 1 \] \[ y = -\frac{3}{7}x + \frac{1}{7} \] From this, we see that the slope \( m \) of the given line is \( -\frac{3}{7} \). The slope of a line perpendicular to this line is the negative reciprocal, which is \( \frac{7}{3} \). Now we can use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), to find the equation of the line we need. With point \( (1, 6) \) and slope \( \frac{7}{3} \): a) \[ y - 6 = \frac{7}{3}(x - 1) \] Simplifying this gives: \[ y - 6 = \frac{7}{3}x - \frac{7}{3} \] \[ y = \frac{7}{3}x + \frac{11}{3} \] Thus, the equation of the line in slope-intercept form is: \[ y = \frac{7}{3}x + \frac{11}{3} \] b) To convert this to standard form ( \( Ax + By = C \) ), we start with the slope-intercept form: \[ y = \frac{7}{3}x + \frac{11}{3} \] Multiply through by 3 to eliminate the fraction: \[ 3y = 7x + 11 \] Rearranging gives: \[ -7x + 3y = 11 \] To write it in standard form with positive \( A \): \[ 7x - 3y = -11 \] Thus, the equation of the line in standard form is: \[ 7x - 3y = -11 \]