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Consider a parallelogram whose vertices are $\mathrm{A}(1,2), \mathrm{B}(4, \mathrm{y}), \mathrm{C}(\mathrm{x}, 6)$ and $\mathrm{D}(3,5)$ taken in order.
What is the value of $\mathrm{AC}^{2}-\mathrm{BD}^{2}$ ?
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What is the value of $\mathrm{AC}^{2}-\mathrm{BD}^{2}$ ?
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The correct answer is:
36
Suppose Mid point of $A C$ and $B D$ is $M(a, b)$.
$a=\frac{1+x}{2}=\frac{3+4}{2}$
$\Rightarrow x=6$
$b=\frac{5+y}{2}=\frac{2+6}{2}$
$\Rightarrow y=3$
$a=\frac{7}{2}, b=4$
$A C^{2}=(1-x)^{2}+(2-6)^{2}=(1-6)^{2}+(-4)^{2}=41$
$B D^{2}=(3-4)^{2}+(5-3)^{2}=1+4=5$
$A C^{2}-B D^{2}=41-5$
$A C^{2}-B D^{2}=36$
$a=\frac{1+x}{2}=\frac{3+4}{2}$
$\Rightarrow x=6$
$b=\frac{5+y}{2}=\frac{2+6}{2}$
$\Rightarrow y=3$
$a=\frac{7}{2}, b=4$
$A C^{2}=(1-x)^{2}+(2-6)^{2}=(1-6)^{2}+(-4)^{2}=41$
$B D^{2}=(3-4)^{2}+(5-3)^{2}=1+4=5$
$A C^{2}-B D^{2}=41-5$
$A C^{2}-B D^{2}=36$
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