Aline segment has endpoint j(2,4) and l(6,8).the point k is the midpoint of jl. what is an equation of a line perpendicular to jl and passing through k

The equation of the line is:
 y-yo = m (x-xo)
 Where,
 m = (y2-y1) / (x2-x1)
 Substituting values:
 m = (8-4) / (6-2)
 m = (4) / (4)
 m = 1
 Then, the equation is:
 y-4 = 1 * (x-2)
 Rewriting:
 y = x-2 + 4
 y = x + 2
 The midpoint is:
 k = ((x1 + x2) / 2, (y1 + y2) / 2)
 k = ((2 + 6) / 2, (4 + 8) / 2)
 k = ((8) / 2, (12) / 2)
 k = (4, 6)
 Then, the equation of the perpendicular line that passes through k is:
 y = -x + b
 Looking for b we have:
 6 = -4 + b
 b = 6 + 4
 b = 10
 Substituting:
 y = -x + 10
 
 An equation of a line perpendicular to jl and passing through k is:
 y = -x + 10

The first answer is correct