Question

# 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 $p$.

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.
$\therefore$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.