Solution 4.1:4b
From Förberedande kurs i matematik 1
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- | {{ | + | If we use the distance formula |
- | < | + | |
- | {{ | + | {{Displayed math||<math>d=\sqrt{(x-a)^2+(y-b)^2}</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>}} |
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} |