If we plot these two points on a graph, we see that it is a straight horizontal line. Slope is found by taking rise/run. Now because the rise is 0, the slope of this line is 0.
Points: (-1, -1) and (-3, 2) Slope: -3/2
2
If: 11x-8y = 32 Then: -8y = -11x+32 And: y = 1.375x-4 in slope-intercept form
|32 - 132| = |-100| = 100 |32 - 132| = |-100| = 100
If you mean: 32-10x+7y = 0 then as a straight line equation it is 7y = 10x-32
It has no slope.
Points: (-1, -1) and (-3, 2) Slope: -3/2
Points: (-1, -1) and (-3, 2) Slope: -3/2
If you mean points: (-3, -5) and (3, 2) then the slope works out as 7/6
Points: (12, 8) and (17, 16) Slope: 8/5 Equation: 5y = 8x-32
32
2
THE QUESTION IS ACTUALLY WORDED. FIND THE EQUATION OF THE LINE THAT CONTAINS THE POINTS P1(-7,-4) AND P2(2,-8). ALGEBRA
30
If you mean points of: (-1, -4) and (3, 2) Slope: 3/2 Equation works out as: 2y = 3x-5
If: 11x-8y = 32 Then: -8y = -11x+32 And: y = 1.375x-4 in slope-intercept form
11x-4y=32