I need help with coordinate geography please! -10 points

ABCD is a rectangle where A is (1,3) and D is (5,5). AC lies on the line 3y=4x+5.
Find
a) the equation of DC
b) the coordinates of C

Now since AC is 3y=4x+5, DC is parallel to the line because it has the same intersecting point "C". And the gradient of AC is 4/3 so gradient of DC is -3/4

Please explain thoroughly plus how do i work out the coordinate of C

sry i tried but didn't get the answer!

First I don't understand how DC is parallel to AC.
Please see if this is correct:
coordinates of A(1,3) and D(5,5) is given and slope of AD=(5-3)/(5-1) =2/4 =1/2
But AD and DC are perpendicular (angles of rectangle 90 deg)
product of slopes= -1
slope of DC = - 1/2
therefore eqn of DC in point-slope form is
y-y1 = m(x-x1)
y-5= -1/2 (x-5)
simplifying: 2y-10 =-x+5 or x+ 2y = 15 --------------eqn i (which is eqn of DC)
b) Also given eqn of AC is 3y = 4x+5 or - 4x +3y = 5 ----------eqn ii
Point C is the point of ontersection of AC and DC. So, on solving eqns i and ii you will get the coordinate
of C.

AC's eqn is 3y=4x+5
DC's slope is -3/4 and its eqn is y-5 = -3/4(x-5) or 3x+4y-35 = 0
the intersection point of AC and DC is obtained by solving 3y=4x+5 and 3x+4y=35
which is (3.4, 6.2)