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What is the perpendicular bisector equation of the line whose end points are at 3 5 and 7 7 on the Cartesian plane?
Wiki User
∙ 6y ago
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
Wiki User
∙ 6y ago
Wiki User
∙ 6y agoIn standard form, it is 2x + y - 16 = 0
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What is the perpendicular bisector equation of the line with end points of -1 4 and 3 8 on the Cartesian plane?
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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 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?
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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 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 equation of a straight line that cuts through the middle of the points of -1 3 and -2 -5 at right angles on the Cartesian plane showing work?
The equation will be a perpendicular bisector equation of the given points:- Points: (-1, 3) and (-2, -5) Midpoint: (-3/2, -1) Slope: 8 Perpendicular slope: -1/8 Perpendicular equation: y--1 = -1/8(x--3/2) => y = -1/8x-3/16-1 Therefore the perpendicular bisector equation is: y = -1/8x -19/16
What is the perpendicular bisector equation to the line segment of -1 -6 and 5 -8?
Points: (-1, -6) and (5, -8) Midpoint: (2, -7) Perpendicular slope: 3 Perpendicular bisector equation: y = 3x -13
What is the perpendicular bisector equation passing between the points 3 -4 and -1 -2?
Points: (3,-4) and (-1, -2) Midpoint: (1,-3) Slope: -1/2 Perpendicular slope: 2 Perpendicular bisector equation in slope intercept form: y = 2x-5
What is the perpendicular bisector equation that meets the line 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 = -13x+10.5
What is the perpendicular bisector equation that meets the line 13 19 and 23 17 at midpoint on the Cartesian plane showing all aspects of work with answer?
Points: (13, 19) and (23, 17) Midpoint: (18, 18) Slope: -1/5 Perpendicular slope: 5 Perpendicular equation: y-18 = 5(x-18) => y = 5x-72
