14. Write the equation of the line that is perpendicular to \( 5 x+3 y=-21 \) and passes through the point \( (-5,1) \).

To find the equation of the line that is perpendicular to \( 5x + 3y = -21 \), we first need to determine the slope of the given line. By rearranging it into slope-intercept form ( \( y = mx + b \) ), we find the slope \( m = -\frac{5}{3} \). Since perpendicular lines have slopes that are negative reciprocals, the slope of the required line will be \( \frac{3}{5} \). Using the point-slope form \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-5, 1) \) and \( m = \frac{3}{5} \), we plug in the values: \[ y - 1 = \frac{3}{5}(x + 5) \] Multiplying out and rearranging gives us: \[ y = \frac{3}{5}x + 4 \] So, the equation of the line is \( y = \frac{3}{5}x + 4 \). Now you have a sparkling new line to work with!