Students of a school are standing in rows and columns in their playground for a drill practice . A, B, C and D are the positions of four students as shown in figure . Is it possible to place Jaspal inn the drill in such a way that he is equidistant from each of the four students A, B C and D ? If so, what should be his position ?

Explanation

Yes , from the figure we observe that the positions of four students A, B , C and D are (3,5),(7,9) , (11,5) and (7,1) respectively i.e., these are four vertices of a a quadrillateral . Now, we find the type of the this quadrilateral . For this , we will all its sides.

Now, AB `=sqrt((7-3)^(2)+(9-5)^(2))`
[ by distance formula , d `=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]`
AB `=sqrt((4)^(2)+(4)^(2))=sqrt(16+16)`
AB`=4sqrt(2)`
BC `=sqrt((11-7)^(2)+(5-9)^(2))=sqrt((4)^(2)+(-4)^(2))`
`=sqrt(16+16)=4sqrt(2)`
and DA `=sqrt((3-7)^(2)+(5-1)^(2))=sqrt((-4)^(2)+(4)^(2))`
`=sqrt(16+16)=4sqrt(2)`
we see that , `AB=BC=CD=DA` i.e., all siedes are equal.
Now , we find lenght of both diagonals .
AC`=sqrt((11-3)^(2)+(5-5)^(2))=sqrt((8)^(2)+0)=8`
and BD `=sqrt((7-7)^(2)+(1-9)^(2))=sqrt(0+(-8)^(2))=8`
Here , AC=BD
Since, `AB=BC=CD=DA and AC=BD`
Which represent a square . Also known the diagonals of a square bisect each other . SO , P be position of jaspal in which he is equidistant from each of the four students A,B, C and D.
`:.` Coordinates of points P = Mid - point of AC
`((3+11)/(2),(5+5)/(2))-=(14/(2),(10)/(2))-=(7,5)`
[since , mid - point of a line segment having points `(x_(1),y_(1))and(x_(2),y_(2))= ((x_(1)+y_(1))/(2),(x_(2)+y_(2))/(2))]`
Hince , the required position of Jaspal is (7,5).