# 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...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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)].

Approved by eNotes Editorial Team