Aline passos through \( (1,1) \) and \( (3,5) \). Ard the the equation of 9 line perpendicular to this ine passing through \( (z,-1) \)

To find the equation of the line perpendicular to the one passing through the points \( (1,1) \) and \( (3,5) \), we first need to determine the slope of the current line. The slope \( m \) is calculated as \( m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 1}{3 - 1} = 2 \). The slope of the perpendicular line is the negative reciprocal, which is \( -\frac{1}{2} \). Now, using the point-slope form \( y - y_1 = m(x - x_1) \) with point \( (z, -1) \) and slope \( -\frac{1}{2} \), the equation becomes \( y + 1 = -\frac{1}{2}(x - z) \). Solving for \( y \) gives us: \[ y = -\frac{1}{2}x + \frac{z}{2} - 1 \] Feel free to replace \( z \) with specific values to find particular equations!