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October 26, 2014

October 26, 2014

Posted by **Owami Tubatse** on Thursday, July 25, 2013 at 12:08pm.

T(2x;y-2) and U(3x+1;y-2)

- mathematics ,analytical ge0metry -
**Steve**, Thursday, July 25, 2013 at 12:13pmJust use the good old Pythagorean Theorem:

d^2 = ∆x^2 + ∆y^2

= ((3x+1)^2-(2x)^2)+((y-2)-(y-2))^2

= (5x^2 + 6x + 1) - (0)

. . .

- mathematics ,analytical ge0metry -
**owami tubatse**, Thursday, July 25, 2013 at 12:36pmStill confused steve can you please elaborate more PLEASE

- mathematics ,analytical ge0metry -
**Reiny**, Thursday, July 25, 2013 at 2:08pmwhat Steve meant to say was:

d^2 = (3x+1 - 2x)^2 + (y-2 - (y-2))^2

= (x+1)^2 + 0^2

d =√(x+1)^2

= x+1

e.g.

let x = 3, y = 5

then T is (6,3) and U is (10,3)

TU = √(4^2 + 0^2)

= √16 = 4

according to my result, d = 3+1 = 4 , as in the test answer.

- mathematics ,analytical ge0metry -
**Steve**, Thursday, July 25, 2013 at 2:37pmouch! Guess I had too many squares working with me!

What was I thinking?

- mathematics ,analytical ge0metry -
**owami tubatse**, Thursday, July 25, 2013 at 10:25pmThank u guys ...I now understand

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