# Given two vertices of a square (2, 4) and (6, 8) is it possible to determine the other two.

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### 1 Answer

In a square all the four sides are of equal length, the diagonals are of equal length and they bisect each other at right angles.

Given the vertices of the square (2, 4) and (6, 8), the other vertices can be uniquely determined if an assumption is made that the these are opposite vertices.

Let the other vertices of the square be (a, b) and (c, d). The slope of the line joining the points (2, 4) and (6, 8) is `(8 - 4)/(6 - 2) = 1` . The length of the diagonal is `sqrt((8-4)^2 +(6-2)^2) = 4*sqrt 2` . The mid point between (2, 4) and (6, 8) is (4, 6). The slope of the line joining each of the points (a, b) and (c, d) and (4, 6) is -1. The distance of each of the points from (4, 6) is `2*sqrt 2` .

`(6 - b)/(4 - a) = -1`

=> 6 - b = a - 4

=> a = 10 - b

`sqrt((4 - a)^2 + (6-b)^2) = 2*sqrt 2`

=> `sqrt((4 - 10 + b)^2 + (6-b)^2) = 2*sqrt 2`

=> `(4 - 10 + b)^2 + (6-b)^2 = 8`

=> `2*(6 - b)^2 = 8`

=> `6 - b = +- 2`

=> b = 8 and b = 4

a = 2 and a = 6

The two values obtained for a and b are the coordinates of both the vertices that we are trying to determine, i.e the coordinates of (c, d) has also been determined.

**The other two vertices of the square are (2, 8) and (6, 4)**