ac + cb = ab = 9
2x - 1 + 3x = 9
5x -1 = 9 So 5x = 10 Thereby x =2. Also ac = 3 and cb = 6
A point on a segment that is halfway between the two endpoints is called the midpoint of the segment. It is the point that divides the segment into two equal parts.
True
If a point lies on a segment whose endpoints are on the sides of an angle but is not an endpoint of the segment, then it is located strictly between the two endpoints of that segment. This means the point is inside the angle formed by the two sides, but not on the angle's boundary itself. The point divides the segment into two smaller segments, both of which lie within the angle.
Yes, if a point is equidistant from the endpoints of a segment, it must be the midpoint of that segment. This is because the midpoint is defined as the point that divides the segment into two equal lengths, making it the only point that maintains equal distance to both endpoints. Therefore, being equidistant from both endpoints confirms that the point is indeed the midpoint.
The point halfway between two endpoints of a line segment is called the midpoint. It can be calculated by averaging the coordinates of the two endpoints. For example, if the endpoints are A(x₁, y₁) and B(x₂, y₂), the midpoint M is given by M((x₁ + x₂)/2, (y₁ + y₂)/2). This point divides the line segment into two equal lengths.
A point on a segment that is halfway between the two endpoints is called the midpoint of the segment. It is the point that divides the segment into two equal parts.
midpoint
True
a middle point
It is the midpoint
It is known as the midpoint of the line segment
If a point lies on a segment whose endpoints are on the sides of an angle but is not an endpoint of the segment, then it is located strictly between the two endpoints of that segment. This means the point is inside the angle formed by the two sides, but not on the angle's boundary itself. The point divides the segment into two smaller segments, both of which lie within the angle.
on the perpendicular bisector of the segment.
Equidistant from the endpoints of the segment.
Yes
then it is equidistant from the endpoints of the segment- apex
on the perpendicular bisector of the segment.