Question

Question 5
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?

Answer 1

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 quadrilateral. Now, we will find the type of this quadrilateral. For this, we will find length of all of its sides.

$Now,AB=\sqrt{(7-3{)}^2+(9-5{)}^2}[bydistanceformula,d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}]AB=\sqrt{(4{)}^2+(4{)}^2}=\sqrt{16+16}AB=4\sqrt{2}BC=\sqrt{(11-7{)}^2+(5-9{)}^2}=\sqrt{(4{)}^2+(-4{)}^2}=\sqrt{16+16}=4\sqrt{2}CD=\sqrt{(7-11{)}^2+(1-5{)}^2}=\sqrt{(-4{)}^2+(-4{)}^2}=\sqrt{16+16}=4\sqrt{2}AndDA=\sqrt{(3-7{)}^2+(5-1{)}^2}=\sqrt{(-4{)}^2+(4{)}^2}=\sqrt{16+16}=4\sqrt{2}$
Now, we can see that AB = BC = CD = DA. i.e., all sides are equal.
Now, we will find length of both the diagonals,
$AC=\sqrt{(11-3{)}^2+(5-5{)}^2}=\sqrt{(8{)}^2+0}=8andBD=\sqrt{(7-7{)}^2+(1-9{)}^2}=\sqrt{0+(-8{)}^2}=8$
Here,AC = BD
Since,AB = BC = CD = DA and AC = BD
The playground represents a square.
Now, the diagonals of a square bisect each other.
So, P (midpoint of diagonal) will be the position of Jaspal so that he is equidistant from each of the four students A, B, C and D.
$\therefore$ Coordinates of point P = Mid-point of AC.
$\equiv (\frac{3+11}{2},\frac{5+5}{2})\equiv (\frac{14}{2},\frac{10}{2})\equiv (7,5)\left[\begin{array}{c}\because Mid-pointoflinesegmenthavingpoints\\ (x_1,y_1)and(x_2,y_2)\\ =(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\end{array}\right]$
Hence, the required position of Jaspal is (7,5)