It is: y--3 = 6(x--1) => y = 6x+3
In standard form: 6x-y+3 = 0
Choose the equation of the line that contains the points (1, -1) and (2, -2).
Points: (1, 2) and (0, -2) Slope: 4 Equation: y = 4x-2
6666
Points: (4, -4) and (-2, 0) Slope: -2/3 Equation: y = -2/3x-4/3 or as 3y = -2x-4
Points: (4, 1) and (5, 2) Slope: 1 Equation: y = x-3 Equation in its general form: x-y-3 = 0
Choose the equation of the line that contains the points (1, -1) and (2, -2).
If you mean points of (-3, 2) and (5, -5) then the equation works out as 8y = -7x-5
Points: (3, -6) and (-3, 0) Slope: -1 Equation: y = -x-3
Points: (3, 0) and (0, -9) Slope: 3 Equation: y = 3x-9
If you mean of points of (3, -4) and (5, 1) then the equation works out as 2y=5x-23
THE QUESTION IS ACTUALLY WORDED. FIND THE EQUATION OF THE LINE THAT CONTAINS THE POINTS P1(-7,-4) AND P2(2,-8). ALGEBRA
Points: (1, 2) and (0, -2) Slope: 4 Equation: y = 4x-2
Points: (8, 10) and (-4, 2) Slope: 2/3 Equation: 3y = 2x+14
Slope: 5 Points: (-2, -3) Equation: y = 5x+7
6666
The equation is (y - 0) = (0 - 2)/(8 - 0)*(x - 8) That is y = (-1/4)*(x - 8) = -x/4 + 2
y = -3x + 5