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If the points $A (a, -11), B (5, b), C (2, 15)$ and $D (1, 1)$ are the vertices of a parallelogram $ABCD$, find the values of $a$ and $b$.
Given:
The points $A (a, -11), B (5, b), C (2, 15)$ and $D (1, 1)$ are the vertices of a parallelogram $ABCD$.
To do:
We have to find the values of $a$ and $b$.
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 \( \left(\frac{2+a}{2}, \frac{-11+15}{2}\right) \)
\( =(\frac{2+a}{2}, \frac{4}{2}) \)
\( =(\frac{2+a}{2}, 2) \)
Similarly,
\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).
The coordinates of \( \mathrm{O} \) are \( \left(\frac{5+1}{2}, \frac{b+1}{2}\right) \)
\( =\left(\frac{6}{2}, \frac{b+1}{2}\right) \)
\( =\left(3, \frac{b+1}{2}\right) \)
On comparing, we get,
\( \frac{2+a}{2}=3 \)
\( \Rightarrow 2+a=6 \)
\( \Rightarrow a=6-2=4 \)
\( \frac{b+1}{2}=2 \)
\( \Rightarrow b+1=4 \)
\( \Rightarrow b=4-1=3 \)
The values of $a$ and $b$ are $4$ and $3$ respectively.
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