Question

If $(7,3),(6,1),(8,2)$ and $(p,4)$ are the vertices of a parallelogram taken in order, then find the value of $p$.

Answer 1

Let the vertices of the parallelogram be $A(7,3),B(6,1),C(8,2)$ and $D(p,4)$
We know that the diagonals of a parallelogram bisect each other.
$\therefore$ The midpoints of the diagonal $AC$ and the diagonal $BD$ coincide.
Hence $(\frac{7+8}{2},\frac{3+2}{2})=(\frac{6+p}{2},\frac{1+4}{2})$
$\Rightarrow (\frac{6+p}{2},\frac{5}{2})=(\frac{15}{2},\frac{15}{2})$
Equating the x-coordinates, we get
$\frac{6+p}{2}=\frac{15}{2}$
$\therefore$ $p=9$