y=-2/3x-4/3
The equation is: y-2 = 2(x-4) => y = 2x-2
y - 2 = 4(x - 7)
Points: (4, -2) and (-5, -2) Slope: (-2--2)/(-5-4) = 0 Equation: y = -2 which will be a staight line parallel to the x axis
If you mean the point of (2, 1) and the line y = 3x+4 Then the perpendicular slope is -1/3 and its equation works out as 3y = -x+5
y=-2/3x-4/3
Coordinate: (1, 2) Slope: 4 Equation: y = 4x-2
what is through (-3 , 2), slope -4 =
The equation is: y-2 = 2(x-4) => y = 2x-2
y - 2 = 4(x - 7)
Points: (4, -4) and (-2, 0) Slope: -2/3 Equation: y = -2/3x-4/3 or as 3y = -2x-4
If you mean: y=3x-4 and the point (2, 1) then the perpendicular equation is 3y=-x+5
y = 4x - 26
Points: (1, 2) and (0, -2) Slope: 4 Equation: y = 4x-2
Points: (4, -2) and (-5, -2) Slope: (-2--2)/(-5-4) = 0 Equation: y = -2 which will be a staight line parallel to the x axis
It is: y = 2x-6
THE QUESTION IS ACTUALLY WORDED. FIND THE EQUATION OF THE LINE THAT CONTAINS THE POINTS P1(-7,-4) AND P2(2,-8). ALGEBRA