What would you like to do?
Is a line segment character of a perpendicular bisect?
A line segment that has been cut exactly in half
Basically the definition of bisect is to separate two parts of a line segment to create two congruent line segments, which leads to them being equal.
â PQR Where PQR form an angle and Q is the angle's vertex. The bisection is the line that goes between the lines QP and QR Bisection is a mathematical tool to find… the root of intervals. Example: â PQR Form an angle of 75Â° A bisection would lead into two smaller angles which can be called â PQA and â RQA, both 37,5Â° And then you can do calculations on the smaller angles, depending on what root you are looking for.
Yes or 'True' ~
No. Since a line is infinite, it has no mid-point. A bisector must go through a midpoint so nothing can bisect a line (not even a segment).
It means it is split in half right down the middle.
It works out as: 2x+y-16 = 0
How do you find the equation of the perpendicular line bisecting the line segment of -2 5 and -8 -3?
if u don't now then i don't now Improved answer as follows:- First find the mid-point of (-2, 5) and (-8, -3) which is (-5, 1) Then find the slope or gradient of (-2, 5) …and (-8, -5) which is 4/3 The perpendicular slope is the negative reciprocal of 4/3 which is -3/4 So the perpendicular bisector passes through (-5, 1) and has a slope of -3/4 Use y -y 1 = m(x -x 1 ) y -1 = -3/4(x- -5) y = -3/4x-11/4 which can expressed in the form of 3x+4y+11 = 0 So the equation of the perpendicular bisector is: 3x+4y+11 = 0
To find its mid point. Which then raises the question why find its midpoint and I cannot answer that.
It bisects the line segment at midpoint at 90 degrees and its slope is the reciprocal of the line segment's slope plus or minus.
Sure. There's even a special name for that line. It's called the "perpendicular bisector" of the segment.
Yes. it is possible to bisect a segment with a perpendicular segment. Follow the link to learn how to do it: http://www.mathopenref.com/constbisectline.html
equidistant from the endpoints of a segment -odewah chin chin