Question

The coordinates of Point S are (2/5, 9 1/8). The coordinates of Point T are (-5 7/10, 9 1/8). What is the distance between Point S and Point T?​

Answer 1

something noteworthy, the y-coordinate for each point is the same, 9⅛, that means is a horizontal line, over which the x-coordinates are at, so since it's a horizontal line, all we need to do is find, what's the distance between
`\bf (2)/(5)\textit{ and }-5(7)/(10)`

of course, let's firstly convert the mixed fraction to improper fraction and then check their difference.

`\bf \stackrel{mixed}{5(7)/(10)}\implies \cfrac{5\cdot 10+7}{10}\implies \stackrel{improper}{\cfrac{57}{10}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{2}{5}-\left[-\cfrac{57}{10} \right]\implies \cfrac{2}{5}+\cfrac{57}{10}\implies \stackrel{\textit{using the LCD of 10}}{\cfrac{(2)2+(1)57}{10}}\implies \cfrac{4+57}{10}\implies \cfrac{61}{10}\implies 6(1)/(10)`

Answer 2

The distance between Point S and Point T is 6.1 unit.

Explanation:

Given : The coordinates of Point S are
`((2)/(5) , 9(1)/(8) )`. The coordinates of Point T are
`(-5(7)/(10),9(1)/(8))`.

To find : What is the distance between Point S and Point T?​

Solution :

The distance formula between two point is

`d=√((x_2-x_1)^2+(y_2-y_1)^2)`

The point S is
`(x_1,y_1)=((2)/(5) , 9(1)/(8) )=((2)/(5) ,(73)/(8) )`

The point T is
`(x_2,y_2)=(-5(7)/(10),9(1)/(8))=(-(57)/(10),(73)/(8))`

Substitute the value,

`d=\sqrt{(-(57)/(10)-(2)/(5))^2+((73)/(8)-(73)/(8))^2}`

`d=\sqrt{((-57-4)/(10))^2+(0)^2}`

`d=\sqrt{((-61)/(10))^2+0}`

`d=(61)/(10)`

`d=6.1`

Therefore, the distance between Point S and Point T is 6.1 unit.