Question

she figures that the dairy section is at point (5, -6). the produce section is at point (-2, -6). the meat section is at point (5, 0). the packaged goods section is at point (-2, 1). the personal hygiene section is at point (0, 1). and the check out counter is at point (-4, 0). which of these pairs of sections are farthest away from one another? a. the dairy section and the meat section b. the meat section and the check out counter c. the produce section and the packaged goods section d. the dairy section and the produce section

Explanation:

Step1: Recall the distance formula

The distance formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.

Step2: Calculate distance for option A

For the dairy section $(5,-6)$ and the meat section $(5,0)$:
$x_1 = 5,y_1=-6,x_2 = 5,y_2 = 0$.
$d_A=\sqrt{(5 - 5)^2+(0+6)^2}=\sqrt{0 + 36}=6$.

Step3: Calculate distance for option B

For the meat section $(5,0)$ and the check - out counter $(-4,0)$:
$x_1 = 5,y_1 = 0,x_2=-4,y_2 = 0$.
$d_B=\sqrt{(-4 - 5)^2+(0 - 0)^2}=\sqrt{(-9)^2+0}=9$.

Step4: Calculate distance for option C

For the produce section $(-2,-6)$ and the packaged goods section $(-2,1)$:
$x_1=-2,y_1=-6,x_2=-2,y_2 = 1$.
$d_C=\sqrt{(-2+2)^2+(1 + 6)^2}=\sqrt{0+49}=7$.

Step5: Calculate distance for option D

For the dairy section $(5,-6)$ and the produce section $(-2,-6)$:
$x_1 = 5,y_1=-6,x_2=-2,y_2=-6$.
$d_D=\sqrt{(-2 - 5)^2+(-6 + 6)^2}=\sqrt{(-7)^2+0}=7$.

Answer:

B. the meat section and the check out counter