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.
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