can a distance be negative in geometry? if so then how would u solve (13,-4) and (13,-6)?

Distance can't be negative.

To find the distance of two points you just use the distance formula:

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

x1, x2, y1, y2 being the x and y values for each of those coordinates.

You can see regardless of which paired coordinates you set as x1, y1 and x2, y2 it will be the same value because of the squaring involved.

Plugging in those values you get

d = sqrt[ (13 - 13)^2 + (-6 - (-4))^2 ]

d = sqrt[ (-6 + 4)^2 ]

d = sqrt[ (-2)^2 ]

d = sqrt[ 4 ]

d = 2

No. Distance can not be negative.

D = √{(x1 - x2)² + (y1 - y2)²}

x1 - x2 = 13 - 13 = 0

y1 - y2 = -4 - 6 = -10

D = √{(0)² + (-10)²[} = √{100} = 10

Always discard the negative of the square root, when computing distance.

let (x1,y1) = (13,-4) and (x2,y2) = (13,-6)

d = [ (x2-x1)^2 + (y2-y1)^2 ] ^/1 2

= [ (13-13)^2 + (-6 - -4 )^2 ]^1/2

= [ 0^2 + (-2)^2 ] ^ 1/2

= [ 4] ^1/2

= 2

The above formula will find the distance between any two points

d = |y2 - y1| = |(-6) - (-4)| = |(-2)| = 2