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Mathematics 15 Online
OpenStudy (anonymous):

Compute the distance between the points (-3, 5) and (4, 6).

OpenStudy (math&ing001):

distance = sqrt((-3-4)²+(5-6)²)

OpenStudy (solomonzelman):

1) Click and hold ALT 2) click the number code (using the numbers that are on the right of the keyboard, and `NOT` the ones below `F1`, `F2`, `F3`, etc., ) 3) release the ALT number code result `2 5 1 ` √ along with some more codes, `0 2 1 5 ` × `2 4 6 ` ÷ `7 5 4 ` ≥ `7 5 5` ≤ `2 4 1 ` or `7 5 3` ± `2 4 7` ≈ just saw the answer and decided to put this up.

OpenStudy (anonymous):

Holy crap!

OpenStudy (anonymous):

@VortexAlliby

OpenStudy (anonymous):

@e.mccormick

OpenStudy (e.mccormick):

The distance formula is related to the Pythagorean Theorem. Know either of those?

OpenStudy (anonymous):

not exactly.

OpenStudy (e.mccormick):

Pythagorean Theorem, for any right triangle with leg lengths a and b and hypotenuse c: \(a^2+b^2=c^2\) Well, when you have two points, you can think of the line between them as the hypotenuse, then the difference in the x values becomes one leg and the difference in the y values becomes the other: Distance Formula, for any two points, \((x_1,y_1)\) and \((x_2,y_2)\) that are distance D appart: \((x_2-x_1)^2+(y_2-y_1)^2=D^2\) or \( D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\) Can you see how the two are related? |dw:1407795973002:dw|

OpenStudy (e.mccormick):

|dw:1407796044948:dw|

OpenStudy (e.mccormick):

See, that is how the two formulas are related. The difference in x becomes the horizontal leg length. The difference in y becomes the vertical leg length. Once you know the length of the legs, you square them, add them, then get the square root. The result is the hypotenuse, which in this case is the distance.

OpenStudy (anonymous):

yes i do but. I am kinda lost because i have to find distance.

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