Question 1: Construct the perpendicular bisector of a line segment with endpoints at (2, 3) and (6, 7).

Answer: Constructed with a compass and ruler, bisecting the segment at its midpoint.

Question 2: Calculate the length of the segment with endpoints at (1, 2) and (5, 6).

Answer: 6

Question 3: Find the coordinates of the midpoint of the segment joining (3, 4) and (7, 8).

Answer: (5, 6)

Question 4: Explain why the perpendicular bisector of a segment passes through its midpoint and is perpendicular to the segment.

Answer: Because the bisector is constructed as the set of points equidistant from both endpoints, it must cross the segment at its midpoint at a right angle.

Question 5: Construct the perpendicular bisector of the segment with endpoints at (4, 2) and (8, 6).

Answer: Constructed with compass and ruler, crossing at the segment's midpoint, perpendicular to it.

Question 6: Given points A(1, 1) and B(5, 5), find the equation of the perpendicular bisector.

Answer: y = -x + 3

Question 7: A point P is equidistant from the points (2, 3) and (6, 7). Show how to locate point P using the perpendicular bisector.

Answer: Construct the perpendicular bisector of the segment joining (2, 3) and (6, 7). Any point on this bisector is equidistant from both points, so locating P on it ensures equidistance.

Question 8: Construct the perpendicular bisector of a segment with length 8 units, with endpoints at (1, 2) and (9, 2).

Answer: A line crossing the segment at (5, 2), perpendicular to it.

Question 9: Find the point that is equidistant from points (3, 4) and (7, 8) and lies on the line y=2x+1.

Answer: Solve the perpendicular bisector of the segment and find its intersection with y=2x+1, which is (5, 11).

Question 10: Identify and correct the mistake in constructing a perpendicular bisector if the bisector does not pass through the segment's midpoint.

Answer: The mistake might be that the compass width was not set to more than half the segment length, or the bisector was drawn without accurate midpoint marking.