Assuming that you mean that those are the (x,y) points, then solve this by using the formula for calculating slope.
Chance in y / chance in x = slope
so,
(-5 - 0) / 0 - 0
Already you can see the problem. The denominator will equal 0, which means that it does not exist.
The slope of that line does not exist, nor does the slope for any vertical line.
On a completely separate note though, the slope of a horizontal line is 0.
Slope = (change_in_y) / (change_in_x) = (4 - 3) / (5 - 0) = 1/5 = 0.2
The slope is normally represented by m so it is 0.
(5, -4) and (0, -3)Slope = ( y2-y1 ) / ( x2-x1 )= (-3 + 4) / (0 - 5) = (1) / (-5)= -0.2
A horizontal line has a slope of 0. If you're using the slope formula, then when the numerator is equal to 0 then the slope is 0.
It is: (5-4)/(2-0) = 1/2
Points: (0, 0) and (-5, -6) Slope: 6/5
If you mean points of (3, 1) and (0, -5) then the slope is 2
Slope = (change_in_y) / (change_in_x) = (4 - 3) / (5 - 0) = 1/5 = 0.2
The slope of points 4 5 and -2 0 would be about 5/6.
this would be a slope of 5
The slope is 0.
If you mean points of: (5, 4) and (0, 3) then the slope is 1/5
Points: (0, 5) and (10, -15) Slope: -2
Points: (1, 5) and (0, 2) Slope: 3
X=5 is a vertical line, so it has no slope. When I say it has no slope, I don't mean the slope is 0, I mean the slope is nonexistent.
There is no slope because it works out as 0
2/5 is the slope