# Answer to Question #8588 in Algebra for brian

Question #8588
How are the distance formula, the Pythagorean Theorem, and equations of circles all related? How does one help you understand the others?
1
2012-04-24T07:42:01-0400
If we place the triangle in the coordinate plane, having A and B coordinates of (x1,y1) and (x2,y2) respectively, it is clear that the length of AC is |x2 - x1| and the length of BC is |x2 - x1|.& We are finding the length, which means that we want a positive value; the absolute value signs guarantee that the result of the operation is always positive. But in the final equation,c^2 = |x2 - x1|&sup2; + |y2-y1|&sup2;, the absolute value sign is not needed since we squared all the terms, and squared numbers are always positive. Getting the square root of both sides we have,

c = &radic;(|x2 - x1|&sup2; + |y2-y1|&sup2;).

We say that c is the distance between A and B, and we call the formula above, the distance formula.

If we want coordinates of B(x,y) where x and y are variables and the distance of B from A constant, say r,& then moving point B about point D maintaining the distance r forms a circle. If A has coordinates (h,k), then
& r&sup2; = (x-h)&sup2; + (y-k)&sup2;

which means that

r = &radic;( (x-h)&sup2; + (y-k)&sup2; ).

Observe that the two& equations above are all of the same form, they are all consequences of the& Pythagorean Theorem.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!