In my Algebraic and Geometric Proof of the Pythagorean Theorem post, we have learned that a right triangle with side lengths and and hypotenuse length , the sum of the squares of and is equal to the square of . Placing it in equation form we have .

If we place the triangle in the coordinate plane, having and coordinates of and respectively, it is clear that the length of is and the length of is . 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,, 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,

We say that is the distance between and , and we call the formula above, the *distance formula*.

If we want coordinates of where and are variables and the distance of from constant, say , then moving point about point maintaining the distance forms a circle. If has coordinates , then which means that .

Observe that the two equations above are all of the same form, they are all consequences of the Pythagorean Theorem. The examples are probably very elementary, but it shows one of the rare beauties of mathematics — the strong connections between and among different concepts.

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Way cool. Did you have students discover these connections using computer software? I would think that may help them “retain” the information and hopefully get “excited” about the “power” of mathematics as a universal language.

Lonnise

Thanks Lonnise. I have tried GeoGebra before. We can show that the rotating point about a fixed point can be constructed using different right triangles. The GeoGebra grid may also help for students to connect the equation of the circle to the distance formula.

hey i really need help with hw could you please help me out!? a2= 6 b2=9 wat is c2

Good article. But the modulus sign within the square root is really not required, as square of any real number is always positive.

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