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(0,8) and (0,-1)

(5,-4) and (1,-7)

2006-09-02 12:31:12 · 2 answers · asked by koalabear 2 in Science & Mathematics Mathematics

2 answers

0 - 0 = 0
-1 - 8 = -7 First point = (0,-7)

5 - 1 = 4
-4 - -7 = -3 Second point = (4,-3)

You have to subtract the X point from the 1st pair with the X point from the 2nd pair. Do the same thing with the Y points. Do this with both problems.

I'm pretty sure this is right, but I'm not sure. Sorry if it's not, I haven't started school yet...

2006-09-02 12:38:37 · answer #1 · answered by Nikki 3 · 0 1

Hmmm, that Pythagorean theorem again?

Difference is P2-P1
Distance D=sqrt( (delta x )^2 + (delta y )^2)

1.(0,8) and (0,-1)
P2-P1 = (0, -9)

delta x = x2-x1= 0 - 0 = 0
delta y = y2-y1 = -1-8 = -9

D=sqrt(0+9^2)=9

2. (5,-4) and (1,-7)
P2-P1 = ((1-5), (-7 –(-4))=
P2-P1 =(-4, -3)

delta x = x2-x1= 1 - 5 = -4
delta y = y2-y1 = -7 +4 = -3

D=sqrt(16+9)=sqrt(25)=5

These methods are useful in vector math.

2006-09-02 12:33:00 · answer #2 · answered by Edward 7 · 0 0

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