Question

the coordinates of the vertices of a rectangle are a (5, 3), b (5, - 9), c (-1, - 9), and d (-1, 3). which measurement is closest to the distance between point b and point d in units? a 5.8 units b 9.1 units c 10.4 units d 13.4 units

Explanation:

Step1: Recall distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For points $B(5,-9)$ and $D(-1,3)$, $x_1 = 5,y_1=-9,x_2=-1,y_2 = 3$.

Step2: Calculate differences

First, find $x_2 - x_1$ and $y_2 - y_1$. $x_2 - x_1=-1 - 5=-6$ and $y_2 - y_1=3-(-9)=3 + 9 = 12$.

Step3: Square the differences

$(x_2 - x_1)^2=(-6)^2 = 36$ and $(y_2 - y_1)^2=12^2 = 144$.

Step4: Sum and square - root

Add the squared differences: $(x_2 - x_1)^2+(y_2 - y_1)^2=36 + 144=180$. Then $d=\sqrt{180}\approx13.4$.

Answer:

D. 13.4 units