Help finding a missing coordinate in a parallelogram!!

The points P(-2,1), Q(-6,4), and R(4,3) are three vertices of a parallelogram PQRS.
Find the coordinate S.

Slope of PQ = (4-1)/(-6+2) = -3/4
Slope of QR = (3-4)/(4+6) = -1/10
Slope of RS = Slope of PQ = -3/4
Slope of PS = Slope of QR = -1/10

RS : (y-3) = -(3/4)(x- 4)
4y-12 = -3x + 12
3x + 4y = 24

PS : (y-1) = -(1/10)(x+2)
10y -10 = -x-2
x+10y = 8

S lies on the intersection of PS and RS:
3(8-10y) + 4y = 24
-30y + 4y = 24 -24 = 0
y = 0
x = 8

S = (8,0)

Or you could recognize that the difference between the coordinates of P and S are the same as the differences between the coordinates of Q and R

S = P + R - Q
Sx = Px + Rx - Qx = -2 + 4+6 = -2+10 = 8
Sy = Py + Ry - Qy = 1 + 3 - 4 = 1 - 1 = 0
S = (Sx, Sy) = (8, 0)

let coordinates of S be (a,b)
let the diagnols of the ||gm(parallelogram) meet at 'o'.
let coordintes of 'o' be (x,y)
in a ||gm diagnols bisect each other,
by mid point theorem,find out co ordinates of 'o'.let p(x1,y1) and r(x2,y2),q(x3,y3)

o(x,y)= x1+x2/2 , y1+y2/2

u ll get....o(x,y)=1,2


nw........o is the midpt of thediagnol QS,so sing mid pt theorem,
o(1,2)= x3+a/2 , y3+b /2 (see above)


the coordinates of S are (8,0).

hewre is the best way to answere this, if one of the lines is diaginal and the others are across and stuff, then just go over the amoun of space the diagonal line takes up (not sure if this makes sense but i hope it does)