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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 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 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 segment of p q and 7p 3q?
In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 0
What is the perpendicular bisector equation of the line joined by the points h k and 3h -5k showing work?
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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 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 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 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 in its general form the perpendicular bisector equation cutting through the line segment of -7 -3 and -1 -4?
Points: (-7, -3) and (-1, -4) Slope: -1/6 Perpendicular slope: 6 Mid-point (-4, -3.5) Equation: y - -3.5 = 6(x - -4) => y = 6x+20.5 Perpendicular bisector equation in its general form: 6x -y+20.5 = 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 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 of the line segment of p q and 7p 3q?
In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 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 of the line joined by the points h k and 3h -5k showing work?
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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?
Points: (s, 2s) and (3s, 8s) Slope: 3 Perpendicular slope: -1/3 Midpoint: (2s, 5s) Equation in its general form: x+3y-17 = 0
