If the points $A (6, 1), B (8, 2), C (9, 4)$ and $D (k, p)$ are the vertices of a parallelogram taken in order, then find the values of $k$ and $p$.

Given:

The points $A (6, 1), B (8, 2), C (9, 4)$ and $D (k, p)$ are the vertices of a parallelogram taken in order.

To do:

We have to find the values of $k$ and $p$.

Solution:

Let the diagonals $AC$ and $BD$ bisect each other at $O$.

Using the mid-point formula, we get,

\( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \)

The coordinates of \( \mathrm{O} \) are \( \left(\frac{6+9}{2}, \frac{1+4}{2}\right) \)

\( =(\frac{15}{2}, \frac{5}{2}) \)

Similarly,

\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).

The coordinates of \( \mathrm{O} \) are \( \left(\frac{8+k}{2}, \frac{2+p}{2}\right) \)

On comparing, we get,

\( \frac{8+k}{2}=\frac{15}{2} \)

\( \Rightarrow 8+k=15 \)

\( \Rightarrow k=15-8=7 \)

\( \frac{2+p}{2}=\frac{5}{2} \)

\( \Rightarrow 2+p=5 \)

\( \Rightarrow p=5-2=3 \)

The values of $k$ and $p$ are $7$ and $3$ respectively.