Question 339394
How do you use the Pythagorean Theorem to find the distance between the pair of points A(0,0), B(6,8) ?
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We want to find the length of the red line below

{{{drawing(400,400,-2,11,-2,11,
locate(0.5,0.7,"A(0,0)"), locate(6.1,8.5,"B(6,8)"),
grid(1), red(line(0,0,6,8)) )}}}


Now we draw a (green) horizontal line segment with
one end at A(0,0) and the other end exactly below
and vertically lined up with B(6,8) (actually the
point C(6,0))  Then we draw a (green) vertical line
segment upward from there C(6,0) to the point B(6,8): 


{{{drawing(400,400,-2,11,-2,11,
locate(0.5,0.7,"A(0,0)"), locate(6.1,8.5,"B(6,8)"),
locate(6.1,.5,"C(6,0)"), 

grid(1), red(line(0,0,6,8)), green(line(0,0,6,0),line(6,0,6,8))


 )}}}

Now we have a right triangle ABC, and we are ready to call on
"ol' 'Thag"

{{{AB^2=AC^2+CB^2}}}

We can see that {{{AC=6}}} and that {{{CB=8}}}, so
upon substituting:

{{{AB^2=6^2+8^2}}}
{{{AB^2=36+64}}}
{{{AB^2=100}}}
{{{AB=sqrt(100)}}}
{{{AB=10}}}

So now we know that AB is 10 units long.

We could convince ourselves that that line really is 10 units
long by taking a compass, placing the sharp point at A(0,0) and
the pencil point at B(6,8), and swinging a big arc and we see
that it cuts both axes at 10.

{{{drawing(400,400,-2,11,-2,11,
locate(0.5,0.7,"A(0,0)"), locate(6.1,8.5,"B(6,8)"),
grid(1), red(line(0,0,6,8)), circle(0,0,10) )}}} 

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FYI, notice that we could have drawn a right triangle this 
way and gotten the same results:

{{{drawing(200,200,-2,11,-2,11,
locate(0.5,0.7,"A(0,0)"), locate(6.1,8.5,"B(6,8)"),
locate(.5,8.1,"C(0,8)"), circle(0,0,10),

grid(1), red(line(0,0,6,8)), green(line(0,0,0,8),line(0,8,6,8))


 )}}}
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Edwin</pre>