A = (s, 2s), B = (3s, 8s)
The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s)
Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3
Gradient of perpendicular to AB = -1/(slope AB) = -1/3
Now,
line through C = (2s, 5s) with gradient -1/3 is
y - 5s = -1/3*(x - 2s) = 1/3*(2s - x)
or 3y - 15s = 2s - x
or x + 3y = 17s
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
It works out in its general form as: x+3y-17s = 0
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
First find the midpoint of the line segment AB which is: (-2, 3) Then find the slope of AB which is: -5/2 The slope of the perpendicular bisector is the positive reciprocal of -5/2 which is 2/5 Then by using the straight line formula of y-y1 = m(x-x1) form an equation for the perpendicular bisector which works out as:- y-3 = 2/5(x-(-2)) y = 2/5x+4/5+3 y = 2/5x+19/5 => 5y = 2x+19 So the equation for the perpendicular bisector can be expressed in the form of:- 2x-5y+19 = 0
The mid-point is needed when the perpendicular bisector equation of a straight line is required. The distance formula is used when the length of a line is required.
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Given a straight line joining the points A and B, the perpendicular bisector is a straight line that passes through the mid-point of AB and is perpendicular to AB.
The perpendicular bisector of a line segment AB is the straight line perpendicular to AB through the midpoint of AB.
A perpendicular bisector is a straight line that divides a side of a triangle in two and is at right angles to that side. An angle bisector is a straight line that divides an angle of a triangle in two.
A compass and a straight edge
It works out in its general form as: x+3y-17s = 0
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
First find the midpoint of the line segment AB which is: (-2, 3) Then find the slope of AB which is: -5/2 The slope of the perpendicular bisector is the positive reciprocal of -5/2 which is 2/5 Then by using the straight line formula of y-y1 = m(x-x1) form an equation for the perpendicular bisector which works out as:- y-3 = 2/5(x-(-2)) y = 2/5x+4/5+3 y = 2/5x+19/5 => 5y = 2x+19 So the equation for the perpendicular bisector can be expressed in the form of:- 2x-5y+19 = 0
The mid-point is needed when the perpendicular bisector equation of a straight line is required. The distance formula is used when the length of a line is required.
The perpendicular bisector of the straight line joining the two points.
First find the mid-point of the line segment which will be the point of intersection of the perpendicular bisector. Then find the slope or gradient of the line segment whose negative reciprocal will be the perpendicular bisector's slope or gradient. Then use y -y1 = m(x -x1) to find the equation of the perpendicular bisector. Mid-point: (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope or gradient: 3q-q/7p-p = 2q/6p = q/3p Slope of perpendicular bisector: -3p/q Equation: y -2q = -3p/q(x -4p) y = -3px/q+12p2/q+2q Multiply all terms by q to eliminate the fractions: qy = -3px+12p2+2q2 Which can be expressed in the form of: 3px+qy-12p2-2q2 = 0
True