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Q: What is the perpendicular bisector equation that meets the line sgment of -1 -6 and 5 -8 on the Cartesian plane showing work?
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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 joining the points of s 2s and 3s 8s on the Cartesian plane showing work?

Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0


What is the perpendicular bisector equation that meets the line segment of 7 3 and -6 1 showing work in addition to the answer?

Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular bisector equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5


What is the length of the line and its equation including its perpendicular bisector equation that spans the points of 2 3 and 5 7 on the Cartesian plane showing key stages of work?

Points: (2, 3) and (5, 7)Length: 5 unitsSlope: 4/3Perpendicular slope: -3/4Midpoint: (3.5, 5)Equation: 3y = 4x+1Bisector equation: 4y = -3x+30.5


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

Related questions

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 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


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 joining the points of s 2s and 3s 8s on the Cartesian plane showing work?

Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0


What is the perpendicular bisector equation that meets the line segment of -2 2 and 6 4 at its midpoint showing work?

Points: (-2, 2) and (6, 4) Midpoint: (2, 3) Slope: 1/4 Perpendicular slope: -4 Perpendicular bisector equation: y-3 = -4(x-2) => y = -4x+11


What is the perpendicular bisector equation that meets the line segment of 7 3 and -6 1 showing work in addition to the answer?

Points: (7, 3) and (-6, 1) Midpoint: (0.5, 2) Slope: 2/13 Perpendicular slope: -13/2 Perpendicular bisector equation: y-2 = -13/2(x-0.5) => 2y = -13x+10.5


What is the length of the line and its equation including its perpendicular bisector equation that spans the points of 2 3 and 5 7 on the Cartesian plane showing key stages of work?

Points: (2, 3) and (5, 7)Length: 5 unitsSlope: 4/3Perpendicular slope: -3/4Midpoint: (3.5, 5)Equation: 3y = 4x+1Bisector equation: 4y = -3x+30.5


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


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 spanning the points of h k and 3h -5k on the Cartesian plane showing key stages of work?

8


What is the equation and its perpendicular bisector equation of the line spanning the points of 3 4 and 0 0 on the Cartesian plane showing work?

Points: (3, 4) and (0, 0)Slope: (4-0)/(3-0) = 4/3Equation: y-4 = 4/3(x-3) => 3y = 4xPerpendicular slope: -3/4Midpoint: (3+0)/2 and (4+0)/2 = (1.5, 2)Bisector equation: y-2 = -3/4(x-1.5) => 4y = -3x+12.5


What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?

8