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How do you form an equation for the perpendicular bisector of the line AB when A is at the point of 7 3 and B is at the point of -6 1?
First find the midpoint of AB which is (1/2, 2)
Then find the slope of AB which is 2/13
The slope of the perpendicular bisector is the negative reciprocal of 2/13 which is -13/2.
Then by using the formula y-y1 = m(x-x1) form an equation for the perpendicular bisector which works out as:-
y -2 = -13/2(x -1/2)
y = -13/2x + 13/4 + 2
y = -13/2x + 21/4
So the equation is: 4y = -26x + 21
What else can I help you with?
What can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
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 the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?
Endpoints: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope of line: 3/1 Slope of perpendicular line: -1/3 Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
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
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 can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
What shape can not form a perpendicular bisector?
A circle cannot form a perpendicular bisector.
What is in its general form the perpendicular bisector equation cutting through the line segment of -7 -3 and -1 -4?
Points: (-7, -3) and (-1, -4) Slope: -1/6 Perpendicular slope: 6 Mid-point (-4, -3.5) Equation: y - -3.5 = 6(x - -4) => y = 6x+20.5 Perpendicular bisector equation in its general form: 6x -y+20.5 = 0
Which of these can not form a perpendicular bisector?
Perpendicular bisector lines intersect at right angles
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 the perpendicular bisector equation passing between the points 3 -4 and -1 -2?
Points: (3,-4) and (-1, -2) Midpoint: (1,-3) Slope: -1/2 Perpendicular slope: 2 Perpendicular bisector equation in slope intercept form: y = 2x-5
What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?
Endpoints: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope of line: 3/1 Slope of perpendicular line: -1/3 Perpendicular bisector 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 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
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
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 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 segment whose endpoints are h k and 3h -5k?
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