Show that the following points are the vertices of a square:
A (6,2), B(2,1), C(1,5) and D(5,6)
Solution
The given points are A (6,2), B(2,1), C(1,5) and D(5,6)
`AB = sqrt((2-6)^2 +(1-2)^2) = sqrt((-4)^2 +(-1)^2) = sqrt(16+1) = sqrt(17) units`
`BC = sqrt((1-2)^2 +(5-1)^2) = sqrt((-1)^2 +(-4)^2) = sqrt(1+16) = sqrt(17) units`
`CD = sqrt((5-1)^2 +(6-5)^2) = sqrt ((4)^2+(1)^2) = sqrt(16+1) = sqrt(17) units`
`DA = sqrt((5-6)^2 +(6-2)^2 )= sqrt((1)^2 +(4)^2) = sqrt(1+16) = sqrt(17) units`
Therefore, AB =BC =CD =DA = 17 units
Also, `AC = sqrt((1-6)^2 +(5-2)^2 ) = sqrt((-5)^2 +(3)^2) = sqrt(25+9) = sqrt(34) units`
`BD = sqrt((5-2)^2 +(6-1)^2) = sqrt((3)^2+(5)^2) = sqrt(9+25) = sqrt(34) units`
Thus, diagonal AC = diagonal BD
Therefore, the given points from a square.
