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If points ${\text{A}}\left( {6,1} \right),{\text{B}}\left( {8,2} \right),{\text{C}}\left( {9,4} \right){\text{ and D}}\left( {p,3} \right)$are the vertices of a parallelogram, taken in order, find the value of $p$.

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Hint- Diagonals of parallelograms are always bisecting each other.

Let ${\text{A}}\left( {6,1} \right),{\text{B}}\left( {8,2} \right),{\text{C}}\left( {9,4} \right){\text{ and D}}\left( {p,3} \right)$be the vertices of a parallelogram ABCD taken in order.
Since the diagonals of a parallelogram bisect each other.
$\therefore $Coordinates of the midpoint of AC$ = $Coordinates of the midpoint of BD
$
   \Rightarrow \left( {\dfrac{{6 + 9}}{2},\dfrac{{1 + 4}}{2}} \right) = \left( {\dfrac{{8 + p}}{2},\dfrac{{2 + 3}}{2}} \right) \\
   \Rightarrow \left( {\dfrac{{15}}{2},\dfrac{5}{2}} \right) = \left( {\dfrac{{8 + p}}{2},\dfrac{5}{2}} \right) \\
   \Rightarrow \dfrac{{15}}{2} = \dfrac{{8 + p}}{2}{\text{ }}\& {\text{ }}\dfrac{5}{2} = \dfrac{5}{2} \\
   \Rightarrow 15 = 8 + p \\
   \Rightarrow p = 15 - 8 = 7 \\
$
Hence the coordinates of the fourth vertex is $\left( {7,3} \right)$
So, the value of $p$ is 7.

Note- In such types of questions always remember the key concept that the diagonals of parallelogram always bisect each other, so it always intersects at midpoint, so apply mid-point property we will get the required answer.
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