find the distance between the following points (2,0)and (5,0) i dont know the formula can u please explain how to solve this problem thanks
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d = [ (y2-y1)^2 + (x2-x1)^2 ]^1/2
Let (x1,y1) = (2,0)
& (x2,y2) = (5,0)
Either point can the first point.
d = [ (y2-y1)^2 + (x2-x1)^2 ]^1/2
= [ (0-0)^2 +( 2-5)^2 ]^1/2
= [ 0 +9]^1/2
= 9^1/2
= 3
Let (x1,y1) = (2,0)
& (x2,y2) = (5,0)
Either point can the first point.
d = [ (y2-y1)^2 + (x2-x1)^2 ]^1/2
= [ (0-0)^2 +( 2-5)^2 ]^1/2
= [ 0 +9]^1/2
= 9^1/2
= 3
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distance equals the square root of (x2-x1)squared+(y2-y1)squared
so you would take 5(x1)-2(x2)squared+0(y2)-0(y1)squared and take the square root of that.
so it would be d= (3)squared+(0)squared
9+0
9
now that you have your number, you take the square root of that, which is 3.
so, your answer is 3.
so you would take 5(x1)-2(x2)squared+0(y2)-0(y1)squared and take the square root of that.
so it would be d= (3)squared+(0)squared
9+0
9
now that you have your number, you take the square root of that, which is 3.
so, your answer is 3.
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since the y-values of the two points are the same, it is basically a horizontal line from x = 2 to x =5
from 2 to 5 there is a distance of 3 units. (5-2)
the distance formula btw is the square root of (x2-x1)^2 - (y2-y1)^2
from 2 to 5 there is a distance of 3 units. (5-2)
the distance formula btw is the square root of (x2-x1)^2 - (y2-y1)^2