Let O be the origin, and P and Q be two point | vedclass

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(C) Let $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ be points on the hyperbola $xy = 12$. The midpoint of $PQ$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) =

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$\frac{7}{2}$

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(C) Let $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ be points on the hyperbola $xy = 12$. The midpoint of $PQ$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) = (\frac{1}{2}, -\frac{1}{2})$. This gives $x_1+x_2 = 1$ and $y_1+y_2 = -1$. Since $y_i = \frac{12}{x_i}$, we have $\frac{12}{x_1} + \frac{12}{x_2} = -1$, which implies $12(x_1+x_2) = -x_1x_2$. Substituting $x_1+x_2 = 1$, we get $12(1) = -x_1x_2$, so $x_1x_2 = -12$. The area of $	riangle OPQ$ with vertices $O(0,0)$, $P(x_1, y_1)$, and $Q(x_2, y_2)$ is given by $\frac{1}{2} |x_1y_2 - x_2y_1|$. Substituting $y_i = \frac{12}{x_i}$, the area is $\frac{1}{2} |x_1(\frac{12}{x_2}) - x_2(\frac{12}{x_1})| = 6 |\frac{x_1^2 - x_2^2}{x_1x_2}| = 6 |\frac{(x_1-x_2)(x_1+x_2)}{x_1x_2}|$. We know $(x_1-x_2)^2 = (x_1+x_2)^2 - 4x_1x_2 = (1)^2 - 4(-12) = 1 + 48 = 49$, so $|x_1-x_2| = 7$. Thus, the area is $6 |\frac{7 	imes 1}{-12}| = 6 	imes \frac{7}{12} = \frac{7}{2}$.

Let $O$ be the origin, and $P$ and $Q$ be two points on the rectangular hyperbola $xy = 12$ such that the midpoint of the line segment $PQ$ is $(\frac{1}{2}, -\frac{1}{2})$. Then the area of the triangle $OPQ$ equals:

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