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What is the perpendicular bisector equation of the line whose coordinates are at h k and 3h -5k on the Cartesian plane showing work?
Wiki User
∙ 9y ago
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Alden Kling
Wiki User
∙ 9y agoPoints: (h, k) and (3h, -5k)
Midpoint: (2h, -2k)
Slope: -3k/h
Perpendicular slope: h/3k
Perpendicular equation:-
y--2k = h/3k(x-2h)
3ky--6k^2 = h(x-h)
3ky--6k^2 = hx-2h^2
3ky = hx-2h^2-6k^2
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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 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 that meets the points of 7 3 and -6 1 on the Cartesian plane?
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-4 = -13x+6.5 => 2y = -13x+10.5 Therefore the perpendicular bisector equation is: 2y = -13x+10.5
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 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 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 with end points of -1 4 and 3 8 on the Cartesian plane?
Points: (-1, 4) and (3, 8) Midpoint (1, 6) Slope: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7
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 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 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 that meets the points of 7 3 and -6 1 on the Cartesian plane?
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-4 = -13x+6.5 => 2y = -13x+10.5 Therefore the perpendicular bisector equation is: 2y = -13x+10.5
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 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 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 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 coordinates are -4 8 and 0 -2 on the Cartesian plane showing work?
Points: (-4, 8) and (0, -2) Slope: (8--2)/((-4-0) = -5/2 Perpendicular slope: 2/5 Midpoint: (-4+0)/2, (8-2)/2 = (-2, 3) Equation: y-3 = 2/5(x--2) Multiply all terms by 5: 5y-15 = 2(x--2) => 5y = 2x+19 Perpendicular bisector equation in its general form: 2x-5y+19 = 0
