Answer:
Step-by-step explanation:
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)
