Question

standard: g.gsr.4.2 - dok2
a quadrilateral has the following vertices: (a(1,2)), (b(2,6)), (c(6,7)), and (d(9,4)). the slope of (overline{ad}) and (overline{bc}) is (\frac{1}{4}).
which statement is true about the quadrilateral?
the mid - point of (overline{ac}) is ((3.5,4.5)) and the mid - point of (overline{bd}) is ((5.5,5)). since the midpoints are not the same, the quadrilateral is a trapezoid.
the mid - point of (overline{ab}) is ((1.5,4)) and the mid - point of (overline{cd}) is ((7.5,5.5)). since the midpoints are not the same, the quadrilateral is a trapezoid.
the mid - point of (overline{ac}) is ((3.5,4.5)) and the mid - point of (overline{bd}) is ((5.5,5)). since the midpoints are not the same, the quadrilateral is a parallelogram.
the mid - point of (overline{ab}) is ((1.5,4)) and the mid - point of (overline{cd}) is ((7.5,5.5)). since the midpoints are not the same, the quadrilateral is a parallelogram.

Explanation:

Step1: Recall mid - point formula

The mid - point formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$.

Step2: Calculate mid - point of $AC$

For points $A(1,2)$ and $C(6,7)$, the mid - point of $\overline{AC}$ is $(\frac{1 + 6}{2},\frac{2+7}{2})=(3.5,4.5)$.

Step3: Calculate mid - point of $BD$

For points $B(2,6)$ and $D(9,4)$, the mid - point of $\overline{BD}$ is $(\frac{2 + 9}{2},\frac{6 + 4}{2})=(5.5,5)$.

Step4: Recall properties of quadrilaterals

In a parallelogram, the diagonals bisect each other (i.e., have the same mid - point). In a trapezoid, the mid - points of the diagonals are different. Since the mid - point of $\overline{AC}$ and $\overline{BD}$ are not the same, and we know that $\overline{AD}\parallel\overline{BC}$ (given the equal slopes), the quadrilateral is a trapezoid.

Answer:

The midpoint of $\overline{AC}$ is $(3.5,4.5)$ and the midpoint of $\overline{BD}$ is $(5.5,5)$. Since the midpoints are not the same, the quadrilateral is a trapezoid.