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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 in 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?
 

Answers (1)

Solution.         Points are A(3, 5), B(7, 9), C(11, 5), D(7, 1)
                   Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}
                   AB= \sqrt{\left ( 7-3 \right )^{2}+\left ( 9-5 \right )^{2}}
                  AB= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}
Length\, of\, BC= \sqrt{\left ( 11-7 \right )^{2}+\left ( 5-9 \right )^{2}}
BC= \sqrt{16+16}
         = \sqrt{32}
         =4 \sqrt{2}
Length\, of\, CD= \sqrt{\left ( 7-11 \right )^{2}+\left ( 1-5 \right )^{2}}
CD= \sqrt{16+16}
         = \sqrt{32}
         =4 \sqrt{2}
Length\, of\, AD= \sqrt{\left ( 3-7 \right )^{2}+\left ( 5-1 \right )^{2}}
AD= \sqrt{16+16}
         = \sqrt{32}
         =4 \sqrt{2}
Length\, of\, AC= \sqrt{\left ( 11-3 \right )^{2}+\left ( 5-5 \right )^{2}}
AC= \sqrt{\left ( 8 \right )}
         = 8
Length\, of\, BD= \sqrt{\left ( 7-7 \right )^{2}+\left ( 1-9 \right )^{2}}
 BD= \sqrt{64}
           = 8
AB = BC = AD, AC = BD
Hence, ABCD is square
The diagonals cut each other at mid-point, which is the equidistance from all four corners of square.
Mid-point \, of \, AC = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )
(x1, y1) =(3, 5)             (x2, y2) = (11, 5)
AC= \left ( \frac{3+11}{2},\frac{5+5}{2} \right )
AC= \left ( 7,5 \right )
This should be the position of Jaspal.

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