Answer:
1) D(1,0), C(-2,-4) or 2) D(9,0), C(6,-4)
Step-by-step explanation:
The vector AB is (2-5, -1-3)= (-3,-4)
The modul of the vector is equal to sqrt (3squared+4squared)=5 (the length of the side AB of square)
Explore the point D (the coordinates of the point is (x,0), y=o, because it is an axis x). AD (x-5, -3)
The modul of AD is sqrt ((x-5)^2+(-3)^2)= sqrt (x^2-10x+25+9), it is equal to the side AD which is equal to AB
sqrt(x^2-10x+34)= 5
x^2-10x+34=25
x^2-10x+9=0
x=1, x=9
D is (1,0) or D is (9,0),
find C, (for D1(1,0))
Find the midpoint of BD (O)
xo= (2+1)/2= 1.5
y0=(-1+0)/2= -0.5
It is the midpoint of Ac too
x0= (xa+xc)/2 1.5 = (5+xc)/2 xc= -2
y0=(ya+yc)/2 -0.5= (3+yc)/2 yc=-4
c(-2,-4)
Find C2 (for D(9,0))
Find the midpoint of BD (O)
x0= (2+9)/2=5.5
y0= (-1+0)/2=-0.5
o(5.5, -0.5)
It is the midpoint of Ac too
x0= (xa+xc)/2 5.5= (5+x)/2 x=6
y0=(ya+yc)/2 -0.5= (3+x)/2 y=-4
