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
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
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.
In its general form of a straight line equation the perpendicular bisector equation works out as:- x-3y+76 = 0
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 perpendicular bisector of the straight line joining the two points.
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
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.
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
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
You have points A, B, and C. Using a compass and straight edge, find a perpendicular bisector of AB (that is, a line that is perpendicular to AB and intersects AB at the midpoint of AB. Next, find a perpendicular bisector of BC. The two lines you found will meet at the center of the circle.
When one draws an isosceles triangle and cast a line straight down from the top, It will result to a perpendicular bisector of the bottom leg. This will only work with an isosceles triangle.
With a compass and a straight edge and the lines must bisect each other at 90 degrees
Maybe, but a straight edge and a pair of compasses would have probably been used to construct a perpendicular line bisector for a given line segment.
A bisector divides an angle into two equal parts. Therefore, if the bisector begins on the middle of a straight line (180 degrees) then the bisector must form a right-angle with the straight line.
Without an equality sign and not knowing the plus or minus values of y and 7 it can't be considered to be a straight line equation therefore finding its perpendicular equation is impossible.
First find the midpoint of (-2, 5) and (-8, -3) which is (-5, 1) Then find the slope of (-2, 5) and (-8, -3) which is 4/3 Slope of the perpendicular bisector is the negative reciprocal of 4/3 which is -3/4 Now form an equation of the straight line with a slope of -3/4 passing through the point (-5, 1) using the formula y-y1 = m(x-x1) The equation works out as: 3x+4y+11 = 0
y = -5x + 9 is the equation of a straight line. It cannot be parallel or perpendicular by itself, you need another line to compare it to.
There are infinitely many lines perpendicular to this line. All of them have the slope of -4/3, if that fact is of any help to you.
First find the midpoint the slope and the perpendicular slope of the points of (p, q) and (7p, 3q) Midpoint = (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope = (3q-q)/(7p-p) = 2q/6p = q/3p Slope of the perpendicular is the negative reciprocal of q/3p which is -3p/q From the above information form an equation for the perpendicular bisector using the straight line formula of y-y1 = m(x-x1) y-2q = -3p/q(x-4p) y-2q = -3px/q+12p2/q y = -3px/q+12p2/q+2q Multiply all terms by q and the perpendicular bisector equation can then be expressed in the form of:- 3px+qy-12p2-2q2 = 0
Any shape with straight edges can have perpendicular edges.
In coordinated geometry the points on a straight line will determine its equation.
Straight line equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Point: (7, 5) Equations intersect: (4, 1) Perpendicular distance: square root of [(7-4)2+(5-1)2] = 5