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Distance Formula

In our straight lines we need to understand the distance formula. The formula may look very odd but it’s just the derivation of a mathematics Pythagoras theorem. We need to prove it. Let make an x line and another y line perpendicular to x line. And they are (2, 1) and another (5,3).we need to find the distance between these two points. There is one method we can, lets name it as (2,1) as Point A and(5,3) as Point B.as we understand this line does not goes straight across ,up and down so we cannot rely on squares and figure out the correct solution to prove it, but we can use geometry from out secondary I and II ,we are aware of Pythagoras theorem. Like if we draw a right triangle and the A and B points are hypotenuse. If we draw a line from point A at 3 to the right, we end up coming below point B and name it as point C.by joining the point B to C.it is a right angle triangle.

We are aware that the distance from point A to C is 3 units and from point C to B is 2 units.by using Pythagoras theorem,a^2+b^2=h^2,applying the units the solution comes as hypotenuse is equal to square root of 13.we can also use a calculator and find the square root of 13 n get the answer in decimals. But by using the same formula, suppose the four bases on a baseball field from a square with 90 ft. sides.

When a player throws the ball from home base to second base, what is the distance of the throw to the nearest tenth? We need to draw it on a sheet of graph mentioning the dimensions and use the Pythagorean Theorem on it. As we know here x2 will be 90 ,x1 is zero.y2 is 90 and y1 is zero so application of formula as Distance is equals to under the root x2 minus x1 to the whole squared plus y2 minus y1 whole square,8100+8100 comes to squared root of 16200.where distance becomes 127.3 feet.to find the distance we can also use the Pythagorean theorem which is nothing but c= under root A square plus B square so by 90^2plus 90^2 it is 16200 square root and c is equals to 127.3 ft. Thus these both ways one can find the distance easily.

There are few more problems solving distance we are given (x1, y1) as 5,-1 and (x2,y2) as 11,7.once we start solving the problem as we already know the distance formula ,we just need to substitute the x and the y values and simplify the problem. Thus finally we get the solution as under the square root 100 and distance is 10.

CR= ∏/4b2s+Vc/Vc

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