The distance between Layana's house and store is 4 units.
Explanation:
Given:
Coordinates of Layana's house = [tex](2\frac{2}{3} , 7\frac{1}{3})[/tex]
Coordinates of the store = [tex](-1\frac{1}{3}, 7\frac{1}{3})[/tex]
It can also be written as:
Coordinates of Layana's house = [tex](\frac{8}{3} , \frac{22}{3})[/tex]
Coordinates of the store = [tex](-\frac{4}{3}, \frac{22}{3})[/tex]
The distance between the house and the shop is:
[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}[/tex]
On substituting the value we get:
[tex]d = \sqrt{(\frac{-4 - 8}{3} )^2 + (\frac{22-22}{3} )^2} \\\\d = \sqrt{\frac{144}{9} } \\\\d = \frac{12}{3} \\\\d = 4 units[/tex]
Therefore, the distance between Layana's house and store is 4 units.
the distance from Layana’s home to the store is 4 units .
Step-by-step explanation:
We know that distance between two points [tex]P(x_1,y_1),Q(x_2,y_2)[/tex] is given by :
⇒ [tex]D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
According to question we have ,
[tex]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})[/tex]
Putting these values in above equation we get :
⇒ [tex]D = \sqrt{(-\frac{4}{3}-\frac{8}{3})^2+(\frac{22}{3}-\frac{22}{3})^2}[/tex]
⇒ [tex]D = \sqrt{(-\frac{12}{3})^2}[/tex]
⇒ [tex]D =\frac{12}{3}[/tex]
⇒ [tex]D =4[/tex]
Therefore , the distance from Layana’s home to the store is 4 units .
