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If $A(6,1)$, $B(8,2)$, $C(9,4)$ and $D(p,3)$ are the vertices of a parallelogram taken in order, what is the value of $p$?
A5
B6
C7
D8
Answer & Solution
Correct answer: C. 7
Principle: diagonals of a parallelogram bisect each other.
1. Midpoint of $AC$ is $\left(\dfrac{6+9}{2},\dfrac{1+4}{2}\right)=\left(\dfrac{15}{2},\dfrac{5}{2}\right)$.
2. Midpoint of $BD$ is $\left(\dfrac{8+p}{2},\dfrac{2+3}{2}\right)=\left(\dfrac{8+p}{2},\dfrac{5}{2}\right)$.
3. Since these midpoints are equal, $\dfrac{15}{2}=\dfrac{8+p}{2}$, so $15=8+p$ and hence $p=7$.
Option D is tempting if one adds the last coordinates incorrectly, while A and B do not make the diagonal midpoints match.
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