A rectangle has side lengths 3 and 4 . One of its vertices is at the point $(1,2)$. Which of the following could not be the coordinates of one of its other vertices?
(A) $(-3,-1)$
(B) $(1,-5)$
(C) $(5,-1)$
(D) $(-2,6)$
(E) $(1,-1)$

## Part C: Each correct answer is worth 8.
Solution 1

Call the given vertex of the rectangle $(1,2)$ point $X$ and name each of the 5 answers to match the letters $A$ through $E$, as shown.

Point $E(1,-1)$ is 3 units below point $X$ (since their $x$-coordinates are equal and their $y$-coordinates differ by 3$)$. Thus, $E(1,-1)$ could be the coordinates of a vertex of a 3 by 4 rectangle having vertex $X$ ( $X$ and $E$ would be adjacent vertices of the rectangle).

Point $C(5,-1)$ is 3 units below and 4 units right of point $X$ (since their $y$-coordinates differ by 3 and their $x$-coordinates differ by 4$)$. Thus, $C(5,-1)$ could be the coordinates of a vertex of a 3 by 4 rectangle having vertex $X$ ( $X$ and $C$ would

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-179.jpg?height=561&width=545&top_left_y=1557&top_left_x=1424)
be opposite vertices of the rectangle).

Point $A(-3,-1)$ is 3 units below and 4 units left of point $X$ (since their $y$-coordinates differ by 3 and their $x$-coordinates differ by 4 ). Thus, $A(-3,-1)$ could be the coordinates of a vertex of a 3 by 4 rectangle having vertex $X$ ( $X$ and $A$ would be opposite vertices of the rectangle).

Point $D(-2,6)$ is 4 units above and 3 units left of point $X$ (since their $y$-coordinates differ by 4 and their $x$-coordinates differ by 3 ). Thus, $D(-2,6)$ could be the coordinates of a vertex of a 3 by 4 rectangle having vertex $X$ ( $X$ and $D$ would be opposite vertices of the rectangle).

The only point remaining is $B(1,-5)$ and since it is possible for each of the other 4 answers to
be one of the other vertices of the rectangle, then it must be $(1,-5)$ that can not be. Point $B(1,-5)$ is 7 units below point $X$ (since their $y$-coordinates differ by 7 ).

How might we show that no two vertices of a 3 by 4 rectangle are 7 units apart?

(See Solution 2).

## Solution 2

The distance between any two adjacent vertices of a 3 by 4 rectangle $P Q R S$ is either 3 or 4 units (such as $P$ and $Q$ or $Q$ and $R$, as shown).

The distance between any two opposite vertices of a rectangle (such as $P$ and $R$ ) can be found using the Pythagorean Theorem.

In the right-angled triangle $P Q R$, we get $P R^{2}=3^{2}+4^{4}$ or $P R^{2}=9+16=25$ and so $P R=\sqrt{25}=5$ (since $P R>0$ ).

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-180.jpg?height=287&width=483&top_left_y=510&top_left_x=1447)

That is, the greatest distance between any two vertices of a 3 by 4 rectangle is 5 units.

As shown and explained in Solution 1, the distance between $X(1,2)$ and $B(1,-5)$ is 7 units. Therefore $(1,-5)$ could not be the coordinates of one of the other vertices of a 3 by 4 rectangle having vertex $X(1,2)$.

ANsweR: (B)