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What is the midpoint of the line segment with endpoints 16 5 and -6 -9?
If you mean endpoints of (16, 5) and (-6, -9) then its midpoint is (5, -2)
Is line segment AB the same as line segment BA?
Yes, while naming a line segment, as long as the two points are on the line, it does not matter what order they are in or which points they are. well their not
What is the class midpoint?
midpoint between 4-16
1The length of the major axis of the ellipse below is 16 and the length of the red line segment is 8 How long is the blue line segment?
8
The class midpoint is?
midpoint between 4-16
Gr equals 16 Br equals 8 and B is between G and R Is B the midpoint of segment Gr?
Yes, because GB = GR - RB
How do you find the midpoint the slope the perpendicular slope and the equation for the perpendicular bisector of the line segment joining the points of 3 5 and 7 7?
Midpoint = (3+7)/2, (5+7)/2 = (5, 6) Slope of line segment = 7-5 divided by 7-3 = 2/4 = 1/2 Slope of the perpendicular = -2 Equation of the perpendicular bisector: y-y1 = m(x-x1) y-6 =-2(x-5) y = -2x+10+6 Equation of the perpendicular bisector is: y = -2x+16
Point b is the midpoint of the line segment pq line segment pq is eight centimeters longer than line segment pb what is the number of centimeters in the length of line segment qb?
Because b is the mid point of pq, pb = qb. pb is half as long as pq Eq#1....pb = 1/2 pq Eq#2....pq = pb +8 Substitute Eq#1 into Eq #2 pq = 1/2 pq + 8 subtracting1/2 pq from both sides 1/2 pq = 8 pq = 16 problem here: you can't subtract 1/2 ... you would have to divide.
What is the perpendicular bisector equation meeting the line segment of 3 5 and 7 7?
Line segment: (3, 5) and (7, 7) Midpoint: (3+7)/2, (5+7)/2 = (5, 6) Slope or gradient: (7-5)/(7-3) = 1/2 Perpendicular slope = -2 Equation: y -6 = -2(x-5) => y = -2x+10+6 => y = -2x+16 So the perpendicular bisector equation is y = -2x+16
What is the midpoint between 6 6 and 16 -6?
Midpoint = (6+16)/2 and (6-6)/2 = (11, 0)
What is the midpoint of the class 16-19?
17.5
What is the perpendicular bisector equation passing through the line segment of 7 7 and 3 5 giving brief details?
Points: (7, 7) and (3, 5) Midpoint: (5, 6) Slope: 1/2 Perpendicular slope: -2 Use: y-6 = -2(x-5) Perpendicular bisector equation: y = -2x+16 or as 2x+y-16 = 0
