If the points A (6, 1), B (8, 2), C (9, 4) and D (k, p)

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (6, 1); B (8, 2); C (9, 4) and D (k, p).

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

In general to find the mid-point $\mathrm{P}(x, y)$ of two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

The mid-point of the diagonals of the parallelogram will coincide.

So,

Co-ordinate of mid-point of $\mathrm{AC}=$ Co-ordinate of mid-point of $\mathrm{BD}$

Therefore,

$\left(\frac{6+9}{2}, \frac{4+1}{2}\right)=\left(\frac{k+8}{2}, \frac{p+2}{2}\right)$

Now equate the individual terms to get the unknown value. So,

$\frac{k+8}{2}=\frac{15}{2}$

$k=7$

Similarly,

$\frac{p+2}{2}=\frac{5}{2}$

$p=3$

Therefore,

$k=7$

$p=3$