The two opposite vertices of a square are (3, 4) and (1, -1). Find the co-ordinates of the other two vertices.

Explanation

Let PQRM be a square and let P(-1, 2) and R(3, 2) be the vertices.
Let the co-ordinates of Q be (x, y).
`because" "PQ=MR`
`rArr" "PQ^(2)=MR^(2)`
`rArr" "(x+1)^(2)+(y-2)^(2)=(x-3)^(2)+(y-2)^(2)`
`" "[because" distance "=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`

`rArr" "x^(2)+1+2x+y^(2)+4-4y=x^(2)+9-6x+y^(2)+4-4y`
`rArr" "2x+1=-6x+9`
`rArr" "8x=8" "rArr" "x=1" "`...(1)
In `DeltaPQR`, we have
`" "PQ^(2)+QR^(2)=PR^(2)`
`" "(x+1)^(2)+(y-2)^(2)+(x-3)^(2)+(y-2)^(2)=(3+1)^(2)+(2-2)^(2)`
`rArr" "x^(2)+1+2x+y^(2)+4-4y+x^(2)+9-6x+y^(2)+4-4y=4^(2)+0^(2)`
`rArr" "2x^(2)+2y^(2)-4x-8y+2=0`
`rArr" "x^(2)+y^(2)-2x-4y+1=0" "...(2)`
Putting x=1 from eq. (1) in eq. (2), we get
`" "1+y^(2)-2-4y+1=0`
`rArr" "y^(2)-4y=0`
`rArr" "y(y-4)=0" "rArr" "y=0or 4`
Hence, the required vertices of square are (1, 0) and (1, 4).