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pythatgoras
\[|a+bi|=\sqrt{a^2+b^2}\]

*pythagoras

in your case \(a=7,b=-1\) so you can just about do it in your head

\[\sqrt{7^2+1^2}=\sqrt{491}=\sqrt{50}=5\sqrt2\]

but what has an "i" to do with a triangle?

|dw:1358396646228:dw|

is the triangle always in that quadrant?

ahh good to know...so is "a" and "b" the length of the lines?

|dw:1358397280980:dw|

|dw:1358397320522:dw|

oh no that is the length of the line....durrr..sorry!!!1

the absolute value of a+bi is the length of the line...correct?