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 $\bf \frac{2}{5}\textit{ and }-5\frac{7}{10}$

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

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

Answer 2

Answer:

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

Step-by-step explanation:

Given : The coordinates of Point S are $(\frac{2}{5} , 9\frac{1}{8} )$. The coordinates of Point T are $(-5\frac{7}{10},9\frac{1}{8})$.

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

Solution :

The distance formula between two point is

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

The point S is $(x_1,y_1)=(\frac{2}{5} , 9\frac{1}{8} )=(\frac{2}{5} ,\frac{73}{8} )$

The point T is $(x_2,y_2)=(-5\frac{7}{10},9\frac{1}{8})=(-\frac{57}{10},\frac{73}{8})$

Substitute the value,

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

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

$d=\sqrt{(\frac{-61}{10})^2+0}$

$d=\frac{61}{10}$

$d=6.1$

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