the easiest way (conceptually, at least), is to translate one of the points to the origin. you can pick either one. suppose you chose to translate (6,2) to the origin. we are thus moving the point (6,2) to (0,0): that is, we are sending (x,y) --> (x-6,y-2). this transformation sends (4,1) to (-2,-1).

so now, we have translating the original problem of reflecting through the line that goes from (6,2) to (4,1), to one of reflecting through the line that goes from (0,0) to (-2,-1).

this is the line y = x/2 (it has slope -1/(-2) = 1/2). in terms of the angle θ this line makes with the x-axis we have:

tanθ = 1/2

cosθ = 1/(√5/2) = 2/√5

sinθ = (1/2)/(√5/2) = 1/√5

so to rotate this line to the x-axis, we have to rotate it -θ, which is multiplying (x,y) (as a column vector) by the matrix:

[ 2/√5 1/√5]

[-1/√5 2/√5], which sends (x,y) to ((2/√5)x + (1/√5)y, -(1/√5)x + (2/√5)y).

now, our line of reflection is just the x-axis, and this just sends (x,y) ---> (x,-y).

finally, we "rotate back", and then "translate back".

so let's put this all together: let's call the translation T, the rotation (by θ) R, and the horizontal reflection H.

the complete transformation is T^-1(R(H(R^-1(T(x,y))))).

T(x,y) = (x-6,y-2). R^-1(T(x,y)) = ((2/√5)(x-6) + (1/√5)(y-2), -(1/√5)(x-6) + (2/√5)(y-2)).

H(R^-1(T(x,y))) = ((2/√5)(x-6) + (1/√5)(y-2), (1/√5)(x-6) - (2/√5)(y-2)).

R(H(R^-1(T(x,y)))) =

[(2/√5)((2/√5)(x-6) + (1/√5)(y-2)) - (1/√5)((1/√5)(x-6) - (2/√5)(y-2)), (1/√5)((2/√5)(x-6) + (1/√5)(y-2)) + (2/√5)((1/√5)(x-6) - (2/√5)(y-2))]

= ((4/5)(x-6) + (2/5)(y-2) - (1/5)(x-6) + (2/5)(y-2), (2/5)(x-6) + (1/5)(y-2) + (2/5)(x-6) - (4/5)(y-2))

= ((3/5)(x-6) + (4/5)(y-2), (4/5)(x-6) - (3/5)(y-2))

finally, we find T^-1(R(H(R^-1(T(x,y))))) = ((3/5)(x-6) + (4/5)(y-2) + 6, (4/5)(x-6) - (3/5)(y-2) + 2)

= (3x/5 + 4y/5 + 4/5, 4x/5 - 3y/5 - 8/5) (you may want to check my arithmetic).