One side of a square is the line from (1,3) to (4,-1). Find the other vertices.

One of the lines of a square connects the points (1,3) and
(4 , -1)

The distance between (1,3) and (4,-1) is sqrt [ (1
- 4)^2 + (3 +1)^2] = sqrt [ 3^2 + 4^2] = sqrt [ 9 + 16] = sqrt 25 =
5

The equation of the line connecting (1,3) and (4,1)
is

y + 1 = [( 3 + 1)/( 1 - 4)] ( x -
4)

=> y + 1 = (4/-3)(x -
4)

=> y + 1 = (-4/3)x -
16/3

=> y = (-4/3)x +
19/3

Therefore the slope of the line is (-4/3) and the y
intercept is 15/3

The sides perpendicular to this side have
a slope 3/4

The equation of the line through (1,3)
is

y - 3 = (3/4)(x -
1)

=> y - 3 = 3x/4 -
3/4

=> y = 3x/4 +
9/4

The equation of the line through ( 4, -1)
is

y +1 = (3/4) ( x -
4)

=> 4y + 4 = 3x -
12

=> 4y - 3x + 16
=0

Now the point (X , Y) on 4y - 3x + 16 =0 which is 5 away
from (4,-1) is given by solving the simultaneous
equations

4Y - 3X + 16 = 0 and sqrt [( X - 4)^2 + (Y +1)^2]
= 5

=> (X - 4)^2 + (3X/4 - 3)^2 =
25

=> X^2 + 16 - 8X + 9X^2/16 + 9 - 9X/2 =
25

=> X1 = 0 , X2 =
8

Y1 = -4 , Y2 = 2

The point
(X , Y) on y = 3x/4 + 9/4 which is 5 away from (1, 3) is given by solving the
simultaneous equations:

4Y = 3X +9 and sqrt [(1 - X)^2 +(
3-Y)^2]=5

=> (1 - X)^2 + (3 - 3X/4 - 9/4)^2 =
25

=> 1 + X^2 - 2X + 9X^2/16 + 9/16 - 9X/8 =
25

X1 = 5 , X2 = -3

Y1 = 6 ,
Y2 = 0

Therefore the two other vertices are
[(-3,0) and (0, -4)] and [(5,6) and (8,2)].