Lösung 4.1:4a
Aus Online Mathematik Brückenkurs 1
K |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
||
| Zeile 17: | Zeile 17: | ||
Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points: | Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points: | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{align} |
d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt] | d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt] | ||
&= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.} | &= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.} | ||
| Zeile 25: | Zeile 25: | ||
Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula | Note: In general, the distance between two points <math>(x,y)</math> and <math>(a,b)</math> is given by the formula | ||
| - | {{ | + | {{Abgesetzte Formel||<math>d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.}</math>}} |
Version vom 08:47, 22. Okt. 2008
If we draw in the points in a coordinate system, we can see the line between the points as the hypotenuse in an imaginary right-angled triangle, where the opposite and adjacent are parallel with the x- and y-axes, respectively.
In this triangle, it is easy to measure the lengths of the opposite and the adjacent, which are simply the distances between the points in the x- and y-directions, respectively.
 |
| ∆x = 5 - 1 = 4 and ∆y = 4 - 1 = 3 |
Using the Pythagorean theorem, we can then calculate the length of the hypotenuse, which is also the distance between the points:
| \displaystyle \begin{align}
d &= \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{4^2+3^2}\\[5pt] &= \sqrt{16+9} = \sqrt{25} = 5\,\textrm{.} \end{align} |
Note: In general, the distance between two points \displaystyle (x,y) and \displaystyle (a,b) is given by the formula
| \displaystyle d = \sqrt{(x-a)^2 + (y-b)^2}\,\textrm{.} |
