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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 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 does a perpendicular bisector do?
A perpendicular bisector [for a given line segment] is a line that meets it at 90 degrees and divides it into two halves.
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 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 that meets the line whose end points are at -2 plus 5 and -8 -3?
Points: (-2, 5) and (-8, -3) Midpoint: (-5, 1) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular bisector equation: y-1 = -3/4(x--5) => 4y-4 = -3x-15 => 4y = -3x-11
What is the perpendicular bisector equation that meets the line segment of -1 -6 and 5 -8 on the Cartesian plane showing work?
Suppose P = (-1, -6) and Q = (5, -8) Then gradient of PQ = (-8 --6)/(5 --1) = -2/6 = -1/3 Therefore gradient of the perp bisector = 3 [because product of gradients = -1] Mid point of PQ = [(-1+5)/2, (-6-8)/2] = (2, -7) Eqn of perp bisector = y + 7 = 3*(x - 2) or 3x - y - 13 = 0
What is prependicular bisector?
A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°.
What is the perpendicular bisector equation that meets the line segment of -1 3 and -2 -5?
Midpoint: (-3/2, -1) Gradient or slope: 8 Perpendicular slope: -1/8 Equation: y- -1 = -1/8(x- -3/2) y = -1/8x -3/16 -1 y = -1/8x -19/16 The perpendicular equation can be expressed in the form of: 2x+16y+19 = 0
What is the perpendicular equation in its general form that meets the line whose coordinates are 2 5 and 11 17 at its midpoint on the Cartesian plane showing all work?
Points: (2, 5) and (11, 17) Midpoint: (6.5, 11) Slope: 4/3 Perpendicular slope: -3/4 Perpendicular equation: y-11 = -3/4(x-6.5) => 4y = -3x+63.5 In its general form: 3x+4y-63.5 = 0
What is the perpendicular distance from the point of 7 and 5 that meets the straight line equation of 3x plus 4y equals 0 on the Cartesian plane showing work?
Equation: 3x+4y = 0 => y = -3/4x Perpendicular slope: 4/3 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (2.08, -1.56) Distance from (7, 5) to (2.08, -1.56) = 8.2 units using the distance formula
