Want this question answered?
Points: (2, 3) and (11, 13) Slope: 10/9 Equation: 9y = 10x+7
Points: (2, 5) and (-4, 1) Slope: 2/3 Equation: 3y = 2x+11
Points: (0, -2) and (20, 42) Slope: 11/5 Equation: 5y = 11x-10 or as y = 2.2x-2.
Points: (-2, 3) and (1, -1) Midpoint: (-0.5, 1) Slope: -4/3 Perpendicular slope: 4/3 Equation: 3y = -4x+1 Perpendicular bisector equation: 4y = 3x+5.5
The equation of a parallel line is of the form 2x - y = c for some c. (-3, -11) is on this lime so 2*(-3) - (-11) = c -6 + 11 = c so that c = 5 and therefore, the equation is 2x - y = 5
Plug both points into the equation of a line, y =m*x + b and then solve the system of equations for m and b to get equation of the line through the points.
Points: (2, 3) and (11, 13) Slope: 10/9 Equation: 9x = 10x+7
Points: (2, 3) and (11, 13) Slope: 10/9 Equation: 9y = 10x+7
Points: (-1, 7) and (-2, 3) Slope: 4 Equation: y = 4x+11
Points: (2, 5) and (-4, 1) Slope: 2/3 Equation: 3y = 2x+11
Points: (0, -2) and (20, 42) Slope: 11/5 Equation: 5y = 11x-10 or as y = 2.2x-2.
(-2, 11)(-3, 14)(2, -1)
Points: (1, 11) and (-2, 2) Slope: (11-2)/(1--2) = 3 Equation: y-11 = 3(x-1) => y = 3x+8 Equation: y-2 = 3(x--2) => y = 3x+8
Points: (7, 0) and (0, 11) Slope: 0-11/7-0 = -11/7 Equation: y-0 = -11/7(x-7) => 7y = -11x+77 Equation: y-11 = -11/7(x-0) => 7y = -11x+77
It is an equation of a straight line.
Y=_10.35x+126.125
Given points: (6, 11), (3, 10)Find: the equation of the line that passes through the given points Solution: First, wee need to find the slope m of the line, and then we can use one of the given points in the point-slope form of the equation of a line. After that you can transform it into the general form of the equation of a line. Let (x1, y1) = (3, 10), and (x2, y2) = (6, 11) slope = m = (y2 - y1)/(x2 - x1) = (11 - 10)/(6 - 3) = 1/3 (y - y1) = m(x - x1)y - 10 = (1/3)(x - 3)y - 10 = (1/3)x - 1y - 10 + 10 - (1/3)x = (1/3)x - (1/3)x + 10 - 1-(1/3)x + y = 9 which is the general form of the required line.