# Finding the distance between pairs of coordinates...

#### jcb33

Soldato
Hey all, maths help I have to find the distance between 2 pairs of coordinates, and I just want to see if im doing it right...

A) (4,1),(7,5)

(7-4)^2 + (5-1)^2

3^2 + 4^2

9+16

25(Square Root)

Associate
Yes.

#### Wile E. Coyote

Associate
Correct, what you have just work out is a 3, 4, 5 triangle

#### Welshy

Soldato
Looks right to me, basically your using Pythagorus theorum.

#### Beenom

Associate
Yes, that's correct. You have to use Pythagoras' theorem (which you did) a^2 + b^2 = c^2.

#### jcb33

Soldato
OP
Thanks all

A) (4,1),(7,5) = 5
B) (3,3),(3,5) = 2
C) (1,7),(10,7) = 9
D) (-3,-6),(3,2) = 10
E) (2,3),(-10,-2) = 13
F) (2,2),(5,5) = (Rounded Down) 4.24

Now to work out the intersection point of two lines

A) x + 2y = 1-, 4x + 3y = 6

Last edited:

#### Welshy

Soldato
jcb33 said:
Thanks all

A) (4,1),(7,5) = 5
B) (3,3),(3,5) = 2
C) (1,7),(10,7) = 9
D) (-3,-6),(3,2) = 10
E) (2,3),(-10,-2) = 13
F) (2,2),(5,5) = (Rounded Down) 4.24

Now to work out the intersection point of two lines

A) x + 2y = 1-, 4x + 3y = 6
If i remember rightly, it has something to do with the number -1?

Permabanned
(3,-2)

#### jcb33

Soldato
OP
X + 2y = 5, 3x - 2y = -1
2y = 5 - x, 2y = -1 -3x
y = 2.5 - 0.5x, y = 0.5 + 1.5x
2.5 - 0.5x = 0.5 + 1.5x

2.5-0.5=1.5x+0.5x
2 = 2x
x = 1

1, 2?

#### touch

Soldato
jcb33 said:
nope, eclp was right (3,-2)

x + 2y = -1
4x + 3y = 6

(4 times the first equation to make X values the same)

4x + 8y = -4
4x + 3y = 6 (top equation minus bottom equation)
=
0x + 5y = -10
5y = -10
y=-2
-----------------
x + 2y = -1 (original equation using y = -2)
x + 2(-2) = -1
x + -4 = -1
x = 3

(3,-2)

#### Akira

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damn nerds clutterin up the fine forums