Question

Layana’s house is located at (2 and two-thirds, 7 and one-third) on a map. The store where she works is located at (–1 and one-third, 7 and one-third). What is the distance from Layana’s home to the store?

Answer 1

the distance from Layana’s home to the store is 4 units .

Step-by-step explanation:

We know that distance between two points $P(x_1,y_1),Q(x_2,y_2)$ is given by :

⇒ $D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

According to question we have ,

$P(x_1,y_1 )= (2\frac{2}{3}, 7\frac{1}{3}) = (\frac{8}{3},\frac{22}{3})\\Q(x_2,y_2) =( -1\frac{1}{3}, 7\frac{1}{3}) = ( -\frac{4}{3}, \frac{22}{3})$

Putting these values in above equation we get :

⇒ $D = \sqrt{(-\frac{4}{3}-\frac{8}{3})^2+(\frac{22}{3}-\frac{22}{3})^2}$

⇒ $D = \sqrt{(-\frac{12}{3})^2}$

⇒ $D =\frac{12}{3}$

⇒ $D =4$

Therefore , the distance from Layana’s home to the store is 4 units .

Answer 2

The distance between Layana's house and store is 4 units.

Explanation:

Given:

Coordinates of Layana's house = $(2\frac{2}{3} , 7\frac{1}{3})$

Coordinates of the store = $(-1\frac{1}{3}, 7\frac{1}{3})$

It can also be written as:

Coordinates of Layana's house = $(\frac{8}{3} , \frac{22}{3})$

Coordinates of the store = $(-\frac{4}{3}, \frac{22}{3})$

The distance between the house and the shop is:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

On substituting the value we get:

$d = \sqrt{(\frac{-4 - 8}{3} )^2 + (\frac{22-22}{3} )^2} \\\\d = \sqrt{\frac{144}{9} } \\\\d = \frac{12}{3} \\\\d = 4 units$

Therefore, the distance between Layana's house and store is 4 units.