Class notes 9/14
September 15, 2007Topics covered in class today included the derivations of the midpoint formula and the distance formula from one dimension to two.
The midpoint formula: it’s a point, so it must have two values, one for the x term and one for the y term, which are separated by a comma, and the point is enclosed within parentheses. Two get the x term, add the x terms of the endpoints of the segment and average them, or divide by two. Do the same thing for the y values.
( (x1 +x2)/2) , (y1 + y2)/2 )
The distance formula is based on the Pythagorean Theorem, which can often be used in its place to determine the length of a segment–all you have to be able to do is determine a right triangle. For the formula, you must determine the difference between the x values, which is like finding the a for the Pythagorean Theorem, and also determine the difference between the y values, which is like finding the b. Square each value, add them together, then take the square root of the sum.
d = sqrt( (x2 – x1) ^2 + (y2 – y1) ^2 )
Because we use this a lot in geometry, you need to memorize and learn this now, instead of waiting until June when you have your final exam.
You must be familiar with simplifying radical numbers also. Only a few combinations of numbers fit the Pythagorean Theorem so that all three numbers are whole numbers. These combinations are called Pythagorean Triples, and here is an address of a list of Pythagorean Triples: http://planetmath.org/encyclopedia/LeastCoprimePythagoreanTriplets.html . You should memorize this list. Or maybe at least the first half dozen on it.
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