answersLogoWhite

0


Best Answer

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

User Avatar

Wiki User

9y ago
This answer is:
User Avatar

Add your answer:

Earn +20 pts
Q: 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?
Write your answer...
Submit
Still have questions?
magnify glass
imp
Related questions

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


The endpoints of a line segment graphed on a Cartesian coordinate system are (-2, 6) and (1, 2) What are the coordinates of the midpoint of the segment?

The endpoints of a line segment graphed on a Cartesian coordinate system are (2, -5) and (-4, 2). What are the coordinates of the midpoint of the segment?


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 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 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 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 coordinates are at h k and 3h -5k on the Cartesian plane showing work?

8


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