Question

which statement proves that pqrs is a parallelogram?
the slopes of \\( \overline { s p } \\) and \\( \overline { r q } \\) are both -2 and \\( s p = r q = \sqrt { 45 } \\).
the slopes of \\( \overline { r s } \\) and \\( \overline { q p } \\) are both 3 and \\( s p = r q = \sqrt { 45 } \\).
the midpoint of \\( \overline { r p } \\) is \\( \left( 4,5 \frac { 1 } { 2 } \
ight) \\) and the slope of \\( \overline { r p } \\) is \\( - \frac { 9 } { 2 } \\).
the midpoint of \\( \overline { s q } \\) is \\( \left( 4,5 \frac { 1 } { 2 } \
ight) \\) and \\( s q = 5 \\).

Explanation:

Step1: Recall the properties of a parallelogram

One of the properties of a parallelogram is that opposite sides are parallel and equal in length.

Step2: Analyze the first option

For the first option, \(SP\) and \(RQ\) are not opposite sides.

Step3: Analyze the second option

For the second option, \(RS\) and \(QP\) are opposite sides.
The slope formula is \(m=\frac{y_2 - y_1}{x_2 - x_1}\).
For \(R(3,10)\) and \(S(2,7)\), the slope of \(\overline{RS}\) is \(m_{RS}=\frac{7 - 10}{2 - 3}=\frac{- 3}{-1}=3\).
For \(Q(6,4)\) and \(P(5,1)\), the slope of \(\overline{QP}\) is \(m_{QP}=\frac{1 - 4}{5 - 6}=\frac{-3}{-1}=3\).
The distance formula is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(S(2,7)\) and \(P(5,1)\), \(SP=\sqrt{(5 - 2)^2+(1 - 7)^2}=\sqrt{9 + 36}=\sqrt{45}\).
For \(R(3,10)\) and \(Q(6,4)\), \(RQ=\sqrt{(6 - 3)^2+(4 - 10)^2}=\sqrt{9+36}=\sqrt{45}\).
Since \(RS\parallel QP\) (because their slopes are equal) and \(SP = RQ\) (because their lengths are equal), this satisfies the property of a parallelogram.

Step4: Analyze the third option

The mid - point and slope of a single diagonal do not prove the figure is a parallelogram.

Step5: Analyze the fourth option

The mid - point and length of a single diagonal do not prove the figure is a parallelogram.

Answer:

The slopes of \(\overline{RS}\) and \(\overline{QP}\) are both \(3\) and \(SP = RQ=\sqrt{45}\).