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What is in its general form the perpendicular bisector equation meeting the line segment of s 2s and 3s 8s at its midpoint?
Wiki User
∙ 12y ago
Points: (s, 2s) and (3s, 8s)
Slope: 3
Perpendicular slope: -1/3
Midpoint: (2s, 5s)
Equation in its general form: x+3y-17 = 0
Wiki User
∙ 12y ago
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What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?
Endpoints: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope of line: 3/1 Slope of perpendicular line: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is the perpendicular bisector equation in its general form that meets the line containing the points 7 3 and -6 1 showing work?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 26x+4y-21 = 0
What is the perpendicular bisector equation of the line segment whose endpoints are h k and 3h -5k?
8
What is the perpendicular bisector equation of the line joined by the points -2 5 and -8 -3 on the Cartesian plane?
Points: (-2, 5) and (-8, -3) Midpoint: (-5, 1) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular equation: y-1 = -3/4(x--5) => 4y = -3x-11 Perpendicular bisector equation in its general form: 3x+4y+11 = 0
What is the perpendicular bisector equation of the line whose end points are at s 2s and 3s 8s on the Cartesian plane?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y -5s = -1/3(x -2s) => 3y = -x +17s Perpendicular bisector equation in its general form: x +3y -17s = 0
What is the perpendicular bisector equation of the line segment whose endpoints are at -4 -10 and 8 -1 on the Cartesian plane?
Endpoints: (-4, -10) and (8, -1) Midpoint: (2, -5.5) Slope: 3/4 Perpendicular slope: -4/3 Perpendicular equation: y --5.5 = -4/3(x-2) => 3y = -4x -8.5 Perpendicular bisector equation in its general form: 4x+3y+8.5 = 0
What is the peroendicular bisector equation of the given points -4 8 and 0 -2 in its general form with work shown?
Points: (-4, 8) and (0, -2) Midpoint: (-2, 3) Slope: -5/2 Perpendicular slope: 2/5 Perpendicular equation: y-3 = 2/5(x--2) => 5y-15 = 2x+4 => 5y = 2x+19 Therefore perpendicular bisector equation in its general form: 2x-5y+19 = 0
What is the perpendicular bisector equation of the line joined by the points of 7 3 and -6 1?
Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Equation: y-2 = -13/2(x-0.5) => 2y-4 = -13(x-0.5) => 2y = -13x+10.5 Perpendicular bisector equation in its general form: 13x+2y -10.5 = 0
What is the perpendicular bisector equation of the line joined by the points of s 2s and 3s 8s?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y-5s = -1/3(x-2s) => 3y-15s = -x+2s => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is the perpendicular bisector equation in its general form of the line whose coodinates are at s 2s and 3s 8s on the Cartesian grid showing key stages of work?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
What is the perpendicular bisector equation of the line segment of 7 3 and -6 1?
Points: (7, 3) and (-6, 1) Midpoint: (1/2, 2) Slope: 2/13 Perpendicular slope: -13/2 Use: y-2 = -13/2(x-1/2) => y = -13/2x+21/4 Perpendicular bisector equation in its general form: 26x+4y-21 = 0
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
8
