What would you like to do?
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.
∠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: ∠PQR Form an angle of 75° A bisection would lead into two smaller angles which can be called ∠PQA and ∠RQA, both 37,5° And then you can do calculations on the smaller angles, depending on what root you are looking for.
Perpendicular line segments are line segments that cross with each other and form angles of 90 degrees.
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 -y1 = m(x -x1) 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