Assuming this is the x-y plane.
Create the right-triangle (-4,3), (-4,-2), (-2,-2)
This is a triangle with sides of 5, 2 and H, the distance between the two endpoints...
using Pythagoras's theorem, we get
H2 = 52 + 22
H2 = 25 + 4
H2 = 29
H = sqrt(29)
This could just as easily have been done with the right triangle (-4,3), (-2,3), (-2,-2)
You can also use the Distance Formula.
It is d2 = (x2 - x1)2 + (y2 - y1)2 where d2 is the distance between the two points squared (you can also say the distance between the two points is the square root of that equation, but as I didn't know how to put the square root sign, I just said d2
In this case, it would be:
d2 = (-4 - (-2))2 + (3 - (-2))2
d2 = (-2)2 + (5)2
d2 = 4 + 25 = 29
d = square root of 29
It doesn't matter which point you use for x1 and y1 as long as they come from the same point. Also the distance formula comes from the Pythagorean Theorem so it's not too important which one you use.
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P1 = (4, -3)P2 = (2, 4)(Distance)2 = (delta-x)2 + (delta-y)2(Distance)2 = (4 - 2)2 + (-3 - 4)2 = (2)2 + (-7)2 = 4 + 49 = 53Distance = sqrt(53) = 7.3 (rounded)
The distance between the points of (4, 3) and (0, 3) is 4 units
(-3-(-6))2 + (7-4)2 = 18 and the square root of this is the distance between the two points
Distance formula: square root of (x1-x2)2+(y1-y2)2
If these are Cartesian coordinates in the standard form (x,y), the distance is 7.28 units, roughly.