Problem 151

Solution 1

The points $A$, $B$, $C$, and $E$ can each be reached from point $P$ by moving 3 units in either the $x$- or $y$-direction and 1 unit in the other direction.

This means that the distance from $P$ to each of these points is $\sqrt{3^2+1^2} = \sqrt{10}$, using the Pythagorean Theorem.

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Therefore, the distance from $P$ to $D$ must be the distance that is different. Solution 2

The coordinates of the 6 points are $A(2, 3)$, $B(4, 5)$, $C(6, 5)$, $D(7, 4)$, $E(8, 1)$, $P(5, 2)$.

Therefore, the distances from $P$ to each of the five other points are $$\begin{aligned} PA & = \sqrt{(5-2)^2+(2-3)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{10} \\ PB & = \sqrt{(5-4)^2+(2-5)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10} \\ PC & = \sqrt{(5-6)^2+(2-5)^2} = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10} \\ PD & = \sqrt{(5-7)^2+(2-4)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{8} \\ PE & = \sqrt{(5-8)^2+(2-1)^2} = \sqrt{(-3)^2 + 1^2} = \sqrt{10}\end{aligned}$$ This tells us that the distance from $P$ to $D$ is the one distance that is different.