Question

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s).$ What is $|r - s|?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

$\dfrac{2}{3}$

$1$

Answer: E

Concepts:transformation perpendicular bisector coordinate geometry

Difficulty rating: 1730

Small Hint:

The rotation center is equidistant from each point and its image, so it lies on the perpendicular bisectors of $AA'$ and $BB'$

Big Hint:

Intersect the perpendicular bisector of $BB'$ (a vertical line) with that of $AA'$

Solution:

A rotation keeps its center equidistant from each point and its image. So $P$ is equidistant from $A$ and $A',$ and from $B$ and $B',$ which puts it at the intersection of two perpendicular bisectors. The bisector of $BB'$ from $(3,3)$ to $(4,3)$ is $x = 3.5.$ The bisector of $AA'$ from $(1,2)$ to $(3,1)$ is $2x - y = 2.5.$ Then $y = 2(3.5) - 2.5 = 4.5,$ so $P = (3.5, 4.5)$ and $|r - s| = |3.5 - 4.5| = 1.$ Thus, E is the correct answer.

2023 AMC 10A Problem 19

Answer: E

Difficulty rating: 1730

Small Hint:

Big Hint:

Solution:

Problem 19 in Other Years