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How do you determine the equation for the perpendicular bisector of the straight line joining the points s 2s and 3s 8s?
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
What else can I help you with?
What is the perpendicular bisector equation of the line y equals 17 -3x that spans the parabola of y equals x squared plus 2x -7?
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
Is a circle can not form a perpendicular bisector?
A circle itself does not form a perpendicular bisector because a perpendicular bisector is a line that divides a segment into two equal parts at a right angle, typically associated with straight segments. However, the concept of a perpendicular bisector can be applied to chords within a circle. The perpendicular bisector of a chord will always pass through the center of the circle.
What is the perpendicular bisector equation of the straight line whose coordinates are s 2s and 3s 8s?
It works out in its general form as: x+3y-17s = 0
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
How do you work out an equation for the perpendicular bisector of the line segment AB when A is at -4 8 and B is at 0 -2?
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
What is the perpendicular bisector equation of the line y equals 17 -3x that spans the parabola of y equals x squared plus 2x -7?
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
What is a characteristic of a perpendicular bisector?
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.
What is a perpendicular bisector of a line segment?
The perpendicular bisector of a line segment AB is the straight line perpendicular to AB through the midpoint of AB.
How is constructing a perpendicular bisector different to constructing an angle bisector?
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.
Is a circle can not form a perpendicular bisector?
A circle itself does not form a perpendicular bisector because a perpendicular bisector is a line that divides a segment into two equal parts at a right angle, typically associated with straight segments. However, the concept of a perpendicular bisector can be applied to chords within a circle. The perpendicular bisector of a chord will always pass through the center of the circle.
Which tools are necessary to construct a perpendicular bisector?
A compass and a straight edge
What is the perpendicular bisector equation of the straight line whose coordinates are s 2s and 3s 8s?
It works out in its general form as: x+3y-17s = 0
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
How do you work out an equation for the perpendicular bisector of the line segment AB when A is at -4 8 and B is at 0 -2?
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
How does midpoint and distance formula fit into the big picture of math?
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.
What is the locus of points equidistant from two points?
The perpendicular bisector of the straight line joining the two points.
How do you work out and find the perpendicular bisector equation meeting the straight line segment of p q and 7p 3q?
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
