find distance of (7,2) and (-5,3)?

Question

help!

Answer 1

the sq. rt. of [ (y2-y1)^2 + (x2-x1)^2 ]

the sq. rt. of [ (3-2)^2 + (-5-7)^2 ]

the sq. rt. of [ (1)^2 + (-12)^2 ]

the sq. rt. of ( 1 + 144)

the sq. rt. of (145)..... which is approximately 12

Answer 2

the distance between (X,Y) and (x,y) is given by

root [ (X - x)^2 + (Y - y)^2 ]

the formula is derived from the pythagoras formula
c^2 = a^2 +b^2

simply connect the dots to form a right angled triangle where the length you want is the hypotenuse (longest side)

and you'll see that (X-x) will be the length of a and (Y-y) will be the length of b

and since your squaring (X-x) and (Y-y) you get a^2 + b^2

but you want c not c^2

so you root a^2 + b^2

which gives you the formula

root [ (X - x)^2 + (Y - y)^2 ]

using this in your case gives you

root [ (7 - (-5))^2 + (2 - 3)^2 ]

root [ (12)^2 + (-1)^2 ]

root [ 144 + 1 ]

so the length is root (145)

Answer 3

the distance formula is x one minus x 2 squared + y one minus y 2 squared (this is all under a square root sign so it should look like this 7 --5= -12 squared = 144 then for y 2-3= -1 squared = 1 so 144 plus 1 equals the square root of 145

Answer 4

use the formula y1 minus y2 over x1 minus x2 so its 2-3/7--5 so its -1/ 12 !!!

Answer 5

use the formula: (x2-x1)2 + (y2-y1)2 ( all this must be written under square root) the 2 outside bracket stands for "square"

Answer 6

and that is 12,04 units