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# If the points $\left( {0,0} \right),\left( {2,0} \right),(0, - 2)$ and $(k, - 2)\;$are concyclic then $k =$A. $2$B. $- 2$C. $0$D. $1$  Hint : Consider the points as coordinates of square

Assume a square ABCD
Since the center of the square is mid point of the diagonal. So it will divide the diagonal in ratio 1:1.
This is a cyclic quadrilateral. In line segment$\,\,AC$, let $O(x,y)$ be midpoint. Then by section formula the ratio between AO:OC = 1:1.
The coordinates of O is $=$ $\left\{ {\left( {\frac{{n{x_1} + m{x_2}}}{{m + n}}} \right),\left( {\dfrac{{n{y_1} + m{y_2}}}{{m + n}}} \right)} \right\}$
$\Rightarrow (x,y) = \left\{ {\left( {\dfrac{{1(0) + 1(0)}}{{1 + 1}}} \right),\left( {\dfrac{{1(0) + 1( - 2)}}{{1 + 1}}} \right)} \right\} \\ \Rightarrow (x,y) = (0, - 1) \\$
∴ Coordinates of $O$ is $(0, - 1)$
Now line segment $BD,O$ is the midpoint
$BO:OD = 1:1$
The coordinates of $O =$$\left\{ {\left( {\dfrac{{n{x_1} + m{x_2}}}{{m + n}}} \right),\left( {\dfrac{{n{y_1} + m{y_2}}}{{m + n}}} \right)} \right\}$

$\Rightarrow (0, - 1) = \left( {\dfrac{{1 \times 2 + 1 \times k}}{{1 + 1}},\dfrac{{1 \times 0 + 1 \times ( - 2)}}{{1 + 1}}} \right) \\ \Rightarrow (0, - 1) = \left( {\dfrac{{2 + k}}{2}, - 1} \right) \\ \\$
From here we can say
$\dfrac{{2 + k}}{2} = 0 \\ k = - 2 \\$
Hence the correct option is B.

Note :- In this question we have considered that those are the coordinates of a square. Now taking two diagonals of a square as we know the center of a square is the midpoint of both the diagonals from this midpoint we know the ratio will be 1:1 by this concept we have solved and got the value of k.

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Table 2 to 20  CBSE Class 11 Maths Chapter 2 - Relations and Functions Formulas  Collinear Points  Table of 20 - Multiplication Table of 20  Factors of 20  Distance Between Two Points  Vector Joining Two Points  Cyclic and Non Cyclic Photophosphorylation  What if the Earth Stopped Spinning?  Square Root of 20  