The diagram shows a rectangle abcd. the point a is (2, 14), b is (−2, 8) and c lies on the x-axis. find
(i) the equation of bc,
(ii) find coordinates of c and d

Since AB and BC are perpendicular so AB.BC=0, (scalar product)
AB=(-4, -6), and BC=(xc+2, yc-8)
AB.BC= -4x-8-6y+48 =0
 so the equation of BC is 2x+3y-20=0 

C is on the line BC, so 2xc+3yc-20=0, but yc=0, so xc=20/2=10

AB and CD are parallel, so det(AB, CD)=0
AB and AD are perpendicular so AB.AD=0 
AB=(-4, -6) and CD =(x-10, y), AD= (x-2, y-14)

det(AB, CD)=0
              x-10-y = 0 equivalent to  x-y = -10
AB.AD=0 
  -4x+8-6y+84=0  equivalent to   2x+3y=44
let's solve

x-y = -10
2x+3y=44
-5y = -64, y = 64/5, x= -10+64/5=14/5

finally  D(14/5, 64/5)

Y’all got to do it all over