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What is the perpendicular bisector equation of the line whose endpoints are s 2s and 3s 8s?
Wiki User
∙ 6y ago
Endpoints: (s, 2s) and (3s, 8s)
Midpoint: (2s, 5s)
Slope: 3
Perpendicular slope: -1/3
Perpendicular bisector equation: 3y = -x+17s or as x+3y-17s = 0
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoIt is x + 3y - 7s = 0.
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What is the perpendicular bisector equation of the line segment whose endpoints are h k and 3h -5k?
8
What is the equation and its perpendicular bisector equation of the line whose end points are at -2 3 and 1 -1 on the Cartesian plane?
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
What is the perpendicular bisector equation of the line segment whose end points are at 3 5 and 7 7?
End points: (3, 5) and (7, 7) Midpoint: (5, 6) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-6 = -2(x-5) => y = -2x+16
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 whose end points are at 3 5 and 7 11 on the Cartesian plane?
Points: (3, 5) and (7, 11) Midpoint: (5, 8) Slope: 3/2 Perpendicular slope: -2/3 Perpendicular equation: y-8=-2/3(x-5) => 3y-24=-2x+10 => 3y=-2x+34 Therefore the perpendicular bisector equation is: 3y = -2x+34
What is the perpendicular bisector equation of the line segment whose endpoints are at 2 9 and 9 2?
Endpoints: (2, 9) and (9, 2) Midpoint: (5.5, 5.5) Slope of line segment: -1 Perpendicular slope: 1 Perpendicular bisector equation: y-5.5 = 1(x-5.5) => y = x
What is the perpendicular bisector equation of the line segment whose endpoints are at -1 3 and -2 -5?
Endpoints: (-1, 3) and (-2, -5) Midpoint: (-3/2, -1) Slope: 8 Perpendicular slope: -1/8 Perpendicular bisector equation: y --1 = -1/8--3/2 => y = -1/8x -19/16
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 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 perpendicular bisector equation of the line segment whose endpoints are at -7 -3 and -1 -4 on the Cartesian plane?
Endpoints: (-7, -3) and (-1, -4) Midpoint: (-4, -3.5) Slope: (-3--4)/(-7--1) = -1/6 Perpendicular slope: 6 Perpendicular bisector equation: y--3.5 = 6(x--4) => y = 6x+20.5
What is the perpendicular bisector equation of the line whose end points are at 3 5 and 7 7 on the Cartesian plane?
Endpoints: (3, 5) and (7,7) Midpoint: (5, 6) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-6 = -2(x-5) => y = -2x+16
What is the equation and its perpendicular bisector equation of the line whose end points are at -2 3 and 1 -1 on the Cartesian plane?
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
What is the perpendicular bisector equation of the line segment whose end points are at 3 5 and 7 7?
End points: (3, 5) and (7, 7) Midpoint: (5, 6) Slope: 1/2 Perpendicular slope: -2 Perpendicular bisector equation: y-6 = -2(x-5) => y = -2x+16
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 end points are at -2 4 and -4 8 on the Cartesian plane?
End points: (-2, 4) and (-4, 8) Midpoint: (-3, 6) Slope: -2 Perpendicular slope: 1/2 Perpendicular bisector equation: y -6 = 1/2(x--3) => y = 0.5x+7.5
What is the perpendicular bisector equation of the straight line whose coordinates are s 2s and 3s 8s?
It works out in its general form as: x+3y-17s = 0
What is the perpendicular bisector equation of the line whose end points are at 3 5 and 7 11 on the Cartesian plane?
Points: (3, 5) and (7, 11) Midpoint: (5, 8) Slope: 3/2 Perpendicular slope: -2/3 Perpendicular equation: y-8=-2/3(x-5) => 3y-24=-2x+10 => 3y=-2x+34 Therefore the perpendicular bisector equation is: 3y = -2x+34
