Yes (in a Euclidean plane)..
It's the theorem that says " One and only one perpendicular can be drawn from a point to a line. "
3y + x = k where k is some constant which can only be determined if a point on it is known. There is no such point given.
True
Yes, but only in principle. In practice, you won't live long enough. Putting it in more positive terms: No matter how many lines have already been drawn perpendicular to a given line [segment], there's always enough room for a lot more of them.
Perpendicular lines will only share one point: the point of intersection, where the two lines meet.
It's the theorem that says " One and only one perpendicular can be drawn from a point to a line. "
Only one
only 1
In a Euclidean plane, only one.
Two points determine a line. Also there is one and only line perpendicular to given line through a given point on the line,. and There is one and only line parallel to given line through a given point not on the line.
Biconditional Statement for: Perpendicular Bisector Theorem: A point is equidistant if and only if the point is on the perpendicular bisector of a segment. Converse of the Perpendicular Bisector Theorem: A point is on the perpendicular bisector of the segment if and only if the point is equidistant from the endpoints of a segment.
3y + x = k where k is some constant which can only be determined if a point on it is known. There is no such point given.
True
It takes 3 non collinear points to define one specific circle. With only two points an infinite number of circles can be drawn. Proof: Given two points A, B draw the line between them. Then find the perpendicular bisector of the line AB. Any point on the perpendicular bisector is equidistant from the two original points, A and B. A circle with center C and radius AC will then pass through points A and B. There are infinite point C's on the perpendicular bisector so there are infinite circles. Given three points A, B and D you can find the perpendicular bisector for line segements AB and then the perpendicular bisector fof line segment BC. The two perpedicular bisectors will not be parallel because the points A, B and D are non collinear. This means the two perpeniducar bisectors will intercept at only one point C(like any two intercepting lines). This point C is equidistant from points A, B, and D. A circle with center C and radius AC will then pass through all three of the points. Since there is only one point C that lies on both perpendicular bisectors, there is only one circle possible.
Yes, but only in principle. In practice, you won't live long enough. Putting it in more positive terms: No matter how many lines have already been drawn perpendicular to a given line [segment], there's always enough room for a lot more of them.
Perpendicular lines will only share one point: the point of intersection, where the two lines meet.
Perpendicular lines intersect at one point only.