find teh distance between (7,2) (-5,3)
2007-12-22 18:31:36 · 6 answers · asked by God Bless America!~ 4 in Science & Mathematics ➔ Mathematics
When finding the distance between two points you subtract the location of x2-x1 and y2-y1=
-5-7=-12
3-2=1
(-12,1)
2007-12-22 18:41:27 · answer #1 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
Given the two points (x1, y1) and (x2, y2), the distance
formula :
distance,d = sqrt[( x2 - x1)^2 + [(y2 - y1)^2]
Therfore, d= sqrt[( -5-7)^2 + ( 3 - 2)^2]
= sqrt[144+ 1]
=sqrt(145 )
=12.04
2007-12-23 02:44:39 · answer #2 · answered by Roslyn** luv maths 2 · 0⤊ 0⤋
D = sqrt((x2 - x1)^2 + (y2 - y1)^2)
D = sqrt((7 - (-5))^2 + (3 - 2)^2)
D = sqrt((7 + 5)^2 + 1^2)
D = sqrt(12^2 + 1)
D = sqrt(144 + 1)
D = sqrt(145)
D = about 12.042
2007-12-23 03:56:32 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
You can use the distance formula between two points.
d = sqrt[(X2-X1)^2 + (Y2-Y1)^2]
d = sqrt[(-5 - 7)^2 + (3 - 2)^2]
d = sqrt[(-12)^2 + (1)^2]
d = 12.04 ANS
teddy boy
2007-12-23 02:42:19 · answer #4 · answered by teddy boy 6 · 0⤊ 0⤋
If you have two points (x1,y1) and (x2,y2), the distance is defined by the distance formula D=sqrt[(x2-x1)^2+(y2-y1)^2]
In this case, x1 is 7, x2 is -5, y1 is 2, and y2 is 3.
x2-x1=-5-7=-12
y2-y1=3-2=1
(x2-x1)^2=(-12)^2=144
(y2-y1)^2=(1)^2=1
D=sqrt[(x2-x1)^2+(y2-y1)^2]
so
D=sqrt(144+1)=sqrt(145)=12.04159
2007-12-23 02:45:48 · answer #5 · answered by retired_dragon 3 · 0⤊ 0⤋
(x2 - x1),(y2 - y1)
-5 - 7= -12
3 - 2= 1
(-12,1)
2007-12-23 02:38:56 · answer #6 · answered by james brown 4 · 0⤊ 0⤋