A (5,3) and B (2,-1) are two verticles of a square ABCD and D is on the x axis. Find the coordinate of C and D​

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

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