Question

standard g.qsr.a.2 - ook2
parallelogram opst has vertices o(0, 10) and s(6, 2). which statement can be used to prove that opst is a rhombus?
the slope of $overline{qs}$ is $-\frac{4}{3}$ and the slope of $overline{rt}$ is $\frac{3}{4}$.
the slope of $overline{qs}$ is $-\frac{4}{3}$ and the slope of $overline{rs}$ is $\frac{3}{4}$.
the length of $overline{qs}$ is 10 and the length of $overline{rt}$ is 10.
the length of $overline{qs}$ is 10 and the length of $overline{rs}$ is 10.

Explanation:

Step1: Recall properties of a rhombus

A parallelogram is a rhombus if its diagonals are perpendicular. The slopes of perpendicular lines are negative - reciprocals of each other.

Step2: Analyze the options

In a parallelogram, the diagonals are the line - segments that connect non - adjacent vertices. For parallelogram \(QPST\), if the slopes of the diagonals are negative reciprocals of each other, then the parallelogram is a rhombus.
The first option gives the slopes of what could be the diagonals of the parallelogram (\(\overline{QS}\) and \(\overline{RT}\)) where the product of the slopes \(-\frac{4}{3}\times\frac{3}{4}=- 1\), indicating that the diagonals are perpendicular.
The second option gives the slope of a side (\(\overline{RS}\)) and a diagonal (\(\overline{QS}\)), which is not relevant for proving the parallelogram is a rhombus.
The third and fourth options give the lengths of a diagonal (\(\overline{QS}\)) and either another diagonal (\(\overline{RT}\)) or a side (\(\overline{RS}\)). Equal lengths of diagonals or a diagonal and a side do not prove a parallelogram is a rhombus.

Answer:

The slope of \(\overline{QS}\) is \(-\frac{4}{3}\) and the slope of \(\overline{RT}\) is \(\frac{3}{4}\)