Pythagorean Theorem: Distance - Pre Algebra

1. A square has vertices at $A(1, 3)$, $B(5, 1)$, $C(3, -3)$, and $D(-1, -1)$. Point $P$ is inside the square such that the sum of the distances from $P$ to all four vertices is minimized. If $P$ has coordinates $(a, b)$, what is the value of $a + b$?
   Answer: $2$ The point that minimizes the sum of distances to all vertices of a polygon is the centroid (center of mass). For a square, this is simply the intersection of the diagonals, which is the average of all vertex coordinates. Centroid: $P = \left(\frac{1+5+3+(-1)}{4}, \frac{3+1+(-3)+(-1)}{4}\right) = \left(\frac{8}{4}, \frac{0}{4}\right) = (2, 0)$. Therefore $a = 2$ and $b = 0$, so $a + b = 2$. Choice A would result from calculation errors in finding the centroid. Choice C might come from using only three vertices. Choice D could result from adding coordinates incorrectly or using a different approach like the circumcenter.
2. A rectangular garden has vertices at $P(2, 1)$, $Q(8, 1)$, $R(8, 5)$, and $S(2, 5)$. A water sprinkler is placed at point $T(5, 7)$. What is the distance from the sprinkler to the closest corner of the garden?
   Answer: $\sqrt{13}$ units We need to find the distance from T(5,7) to each corner: to P(2,1): $\sqrt{(5-2)^2 + (7-1)^2} = \sqrt{9+36} = \sqrt{45} = 3\sqrt{5}$; to Q(8,1): $\sqrt{(5-8)^2 + (7-1)^2} = \sqrt{9+36} = 3\sqrt{5}$; to R(8,5): $\sqrt{(5-8)^2 + (7-5)^2} = \sqrt{9+4} = \sqrt{13}$; to S(2,5): $\sqrt{(5-2)^2 + (7-5)^2} = \sqrt{9+4} = \sqrt{13}$. The smallest distance is $\sqrt{13}$. Choice B equals $\sqrt{20}$ which is larger than $\sqrt{13}$. Choice C equals $\sqrt{17}$ which is also larger. Choice D equals $\sqrt{18}$ which is larger than $\sqrt{13}$.

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