Question #8588

How are the distance formula, the Pythagorean Theorem, and equations of circles all related? How does one help you understand the others?

Expert's answer

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|² + |y2-y1|², 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 = √(|x2 - x1|² + |y2-y1|²).

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

& r² = (x-h)² + (y-k)²

which means that

r = √( (x-h)² + (y-k)² ).

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

c = √(|x2 - x1|² + |y2-y1|²).

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

& r² = (x-h)² + (y-k)²

which means that

r = √( (x-h)² + (y-k)² ).

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

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