In the diagram, which of the following points is at a different distance from $P$ than the rest of the points?
(A) $A$
(B) $B$
(C) $C$
(D) $D$
(E) $E$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-015.jpg?height=458&width=463&top_left_y=920&top_left_x=1357)
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.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-028.jpg?height=461&width=464&top_left_y=1973&top_left_x=901)

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

$$
\begin{aligned}
& P A=\sqrt{(5-2)^{2}+(2-3)^{2}}=\sqrt{3^{2}+(-1)^{2}}=\sqrt{10} \\
& P B=\sqrt{(5-4)^{2}+(2-5)^{2}}=\sqrt{1^{2}+(-3)^{2}}=\sqrt{10} \\
& P C=\sqrt{(5-6)^{2}+(2-5)^{2}}=\sqrt{(-1)^{2}+(-3)^{2}}=\sqrt{10} \\
& P D=\sqrt{(5-7)^{2}+(2-4)^{2}}=\sqrt{(-2)^{2}+(-2)^{2}}=\sqrt{8} \\
& P E=\sqrt{(5-8)^{2}+(2-1)^{2}}=\sqrt{(-3)^{2}+1^{2}}=\sqrt{10}
\end{aligned}
$$

This tells us that the distance from $P$ to $D$ is the one distance that is different.

ANswer: (D)