Question

which statement proves that the diagonals of square pqrs are perpendicular bisectors of each other?
the length of sp, pq, rq, and sr are each 5.
the slope of sp and rq is -\frac{4}{3} and the slope of sr and pq is \frac{3}{4}.
the length of sq and rp are both \sqrt{50}.
the midpoint of both diagonals is (4\frac{1}{2}, 5\frac{1}{2}), the slope of rp is 7, and the slope of sq is -\frac{1}{7}.

Explanation:

Step1: Recall perpendicular - bisector properties

Two lines are perpendicular if the product of their slopes is - 1, and they are bisectors of each other if they have the same mid - point.

Step2: Analyze each option

• Option 1: The lengths of the sides ($SP$, $PQ$, $RQ$, $SR$) being equal proves it's a rhombus, not that the diagonals are perpendicular bisectors.
• Option 2: The slopes of the sides ($SP$, $RQ$, $SR$, $PQ$) are given, not the diagonals.
• Option 3: The lengths of the diagonals ($SQ$ and $RP$) being equal does not prove they are perpendicular bisectors.
• Option 4: The mid - point of both diagonals is the same ($(4.5,5.5)$), which means they bisect each other. Also, the product of the slopes of $\overline{RP}$ ($m_{RP}=7$) and $\overline{SQ}$ ($m_{SQ}=-\frac{1}{7}$) is $7\times(-\frac{1}{7})=- 1$, so they are perpendicular.

Answer:

The mid - point of both diagonals is $(4\frac{1}{2},5\frac{1}{2})$, the slope of $\overline{RP}$ is 7, and the slope of $\overline{SQ}$ is $-\frac{1}{7}$.