Solution 4.1:4b
From Förberedande kurs i matematik 1
(Difference between revisions)
m |
|||
Line 1: | Line 1: | ||
If we use the distance formula | If we use the distance formula | ||
+ | {{Displayed math||<math>d=\sqrt{(x-a)^2+(y-b)^2}</math>}} | ||
- | <math> | + | to determine the distance between the points <math>(x,y) = (-2,5)</math> and <math>(a,b) = (3,-1)</math>, we get |
- | + | {{Displayed math||<math>\begin{align} | |
- | + | d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt] | |
- | + | &= \sqrt{(-5)^2+6^2}\\[5pt] | |
- | + | &= \sqrt{25+36}\\[5pt] | |
- | + | &= \sqrt{61}\,\textrm{.} | |
- | + | \end{align}</math>}} | |
- | + | ||
- | <math>\begin{align} | + | |
- | & | + | |
- | & =\sqrt{ | + | |
- | \end{align}</math> | + |
Current revision
If we use the distance formula
\displaystyle d=\sqrt{(x-a)^2+(y-b)^2} |
to determine the distance between the points \displaystyle (x,y) = (-2,5) and \displaystyle (a,b) = (3,-1), we get
\displaystyle \begin{align}
d &= \sqrt{(-2-3)^2+(5-(-1))^2}\\[5pt] &= \sqrt{(-5)^2+6^2}\\[5pt] &= \sqrt{25+36}\\[5pt] &= \sqrt{61}\,\textrm{.} \end{align} |