how to solve the distance between two points?

Question

please answer this..thank you..!!!

Answer 1

If d is the distance between two points (x1, y1) and (x2, y2), the relationship is given by the equation:

d^2 = (x2 - x1)^2 + (y2 - y1)^2

Answer 2

how to solve the distance between two points?

answer:

it really depends upon the given. but in mathematics one way we can solve distance between two points is by using Pythagorean Theorem which is

c = square root of (a^2 + b^2)

where in c is the hypotenuse of the right triangle or
the resultant (longest side)
a is the altitude or the y-distance (y2-y1) and
b is the base or the x-distance (x2-x1).

enjoy solving:-)

Answer 3

that's a sq. root of the sum of the squares project. the finished x distance is 7 -(-3) = 10. the finished y distance is 4 - 0 = 4. Pythagorean formulation: sqrt( x^2 + y^2 ) or sqrt(a hundred +sixteen ) = 10.seventy seven

Answer 4

It actually depends on whether you're on a plane or some other surface such as a sphere, and although you'll get the SAME result no matter which coordinate system you are working with, you have to convert to rectangular (Cartesian) coordinates and use the the Pythagorean Theorem which has been correctly stated above and which can be extended to more than just 2-D.

Answer 5

Usually Pythagoras' Theorem although strictly this is for displacement as opposed to distance but it usually works

D = sqrt[(x2-x1)^2+(y2-y1)^2]

or

sqrt[(x2-x1)^2+(y2-y1)^2+
(z2-z1)^2]

The z term is meant to be there but YA don't like longish horizontal text <_<
use the first equation for 2D, the second for 3D

Answer 6

For this you should be given 2 points ex. (2,1), (3,2)
(x1,y1),(x2,y2)

now the formula- (x2-x1)2+(y2-y1)2 then get the square root of your final answer

Note the 2 outside the bracket means to square the contents of the bracket

Answer 7

distance between 2 points depends on following
1] real line.
2] 2-dimensional eucledian space.
3] 3-dim.. euclid space
4] 4- dim.. euclid space.
.................... n- dim.. euclid space.

now, 1] real line
let the points be a and b.
distance = l a-bl i.e [ absolut value of a-b]
2] 2-d space
let the points be (a,b) and (c,d)

distance = [ (a-c)^2 + (b-d)^2]^.5
3] 3-d space
the points be (a,b,c) and (p,q,r)

distance = [ (a-p)^2 + (b-q)^2 + (c-r)^2 ]^.5
........etc then you can generalise the formula for n tuples.

Answer 8

D= {(x2-x1)+(y2-y1)}^-1/2
when two cordinate points were given.
well i'm not getting your question

Answer 9

square root of (y2 - y1)^2 + (x2 - x1)^2

Answer 10

use a ruler, duh