what is the approximate distance from (4.1) to (7,8)

Question

anonymous · Answer

Use the formula:

$$Distance = \sqrt{(x_1-x_2)^2 + (y_1 - y_2)^2}$$

anonymous · Answer

First step:

change in x, the horizontal change: $\Delta x = x_2-x_1$
change in y, the vertical change:      $\Delta y = y_2-y_1$

next...

anonymous · Answer

Here,
$$x_1 = 4, \; x_2 = 7, \; y_1 = 1, \; y_2 = 8$$

Plug these values in the formula..

anonymous · Answer

is it 5 6 7 or 8 units?

anonymous · Answer

Find it yourself we have provided you the hints and try it once...

anonymous · Answer

The second step is to use the Pythagorean Theorem!

a^2 + b^2 = c^2

In this case "c" is your distance, they're the same thing.

So replacing "c" with "d", replacing "a" with $\Delta x$ and "b" with $\Delta y$ and solving:

$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

|dw:1342809102034:dw|

@waterineyes that works fine is it's directionless (as this type of problem is) but you're going to confuse the poor person later if they get to physics and start using vectors :-)   Better to always define change and final minus initial, in whatever order you were given.