# 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. $1$

Answer

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366.3k+ views

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.

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.

Last updated date: 28th Sep 2023

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Total views: 366.3k

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