Question

practice 5 (from unit 1, lesson 3)
here are 2 points in the plane.

a. using only a straightedge, can you find points in the plane that are the same distance from points a and b?
a yes
b no
explain your reasoning.

b. using only a compass, can you find points in the plane that are the same distance from points a and b?

Explanation:

Part a

To find points equidistant from \( A \) and \( B \), we need the perpendicular bisector of segment \( AB \). A straightedge is used for drawing straight lines (like \( AB \) or other lines), but to find the set of points equidistant from \( A \) and \( B \), we need the perpendicular bisector. A straightedge alone can't construct the perpendicular bisector (which requires a compass for the arcs to find the midpoint and the perpendicular line). However, wait—actually, the set of points equidistant from \( A \) and \( B \) is the perpendicular bisector of \( AB \). But can we find some points on it with a straightedge? Wait, no—because to find equidistant points, the key is that the perpendicular bisector is the locus. But with a straightedge, we can draw the line segment \( AB \), and the midpoint of \( AB \) is on the perpendicular bisector (so it's equidistant from \( A \) and \( B \)). Wait, to find the midpoint with a straightedge? No, the midpoint can be found by drawing two lines from \( A \) and \( B \) to form a parallelogram? Wait, no—actually, the set of points equidistant from \( A \) and \( B \) is the perpendicular bisector. But with a straightedge, we can draw the line \( AB \), and any point on the perpendicular bisector is equidistant. But to find the perpendicular bisector, we need a compass. Wait, but the question is "can you find points" (not necessarily all, just some). Wait, the midpoint of \( AB \) is equidistant from \( A \) and \( B \), and we can find the midpoint of \( AB \) using a straightedge by drawing two lines from \( A \) and \( B \) to form a parallelogram? No, that's not right. Wait, actually, the perpendicular bisector of \( AB \) is the set of all points equidistant from \( A \) and \( B \). To find a point on it, the midpoint is on it. But to find the midpoint with a straightedge, we can use the method of drawing two lines from \( A \) and \( B \) to intersect at two points, forming a parallelogram, and the diagonals bisect each other. Wait, maybe. But actually, the correct reasoning is: The set of points equidistant from \( A \) and \( B \) is the perpendicular bisector of \( \overline{AB} \). A straightedge can be used to draw \( \overline{AB} \), and then, by constructing a parallelogram with \( A \) and \( B \) as vertices, the diagonals of the parallelogram bisect each other, so the intersection of the diagonals is the midpoint of \( \overline{AB} \) (which is on the perpendicular bisector, hence equidistant from \( A \) and \( B \)). Also, any line perpendicular to \( \overline{AB} \) at its midpoint is the perpendicular bisector. But with a straightedge, we can draw lines. Wait, maybe the answer is "Yes" because the midpoint of \( AB \) is equidistant from \( A \) and \( B \), and we can find the midpoint using a straightedge (by drawing two lines from \( A \) and \( B \) to form a parallelogram, as the diagonals of a parallelogram bisect each other). Alternatively, the perpendicular bisector is the set of points, and with a straightedge, we can draw the line \( AB \), and then draw a line perpendicular to \( AB \) at its midpoint (but to draw the perpendicular, we need a compass). Wait, I think I made a mistake. Let's recall: The perpendicular bisector of a segment is the set of all points equidistant from the two endpoints. To construct the perpendicular bisector, we need a compass (to draw arcs from \( A \) and \( B \) with the same radius). A straightedge alone can't construct the perpendicular bisector, but can we find some points equidistant? The midpoint is equidistant, and to find t…

To find points equidistant from \( A \) and \( B \), we consider the locus (set of all such points) which is the perpendicular bisector of \( \overline{AB} \). A compass can be used to draw circles (or arcs) centered at \( A \) and \( B \) with the same radius. The intersection points of these two circles (arcs) lie on the perpendicular bisector of \( \overline{AB} \), meaning they are equidistant from \( A \) and \( B \). By adjusting the radius of the compass (as long as it's greater than half the length of \( \overline{AB} \)), we can find multiple points (the intersections of the two circles) that are equidistant from \( A \) and \( B \). So using only a compass, we can find such points.

Answer:

A. Yes

Part b