Question

here are 3 points in the plane. select all the straightedge and compass constructions needed to locate the point that is the same distance from all 3 points. a construct the bisector of angle zxy. b construct the bisector of angle zxy. c construct the perpendicular bisector of yz. d construct the perpendicular bisector of xy. e construct a line perpendicular to xy through point z. f construct a line perpendicular to yz through point x.

Explanation:

Step1: Recall the property of circum - center

The point that is equidistant from three non - collinear points in a plane is the circum - center of the triangle formed by those three points.

Step2: Recall the construction of circum - center

The circum - center is the intersection of the perpendicular bisectors of the sides of the triangle. So, to find the point that is equidistant from points \(X\), \(Y\), and \(Z\), we need to construct the perpendicular bisectors of the sides of \(\triangle XYZ\).

Answer:

C. Construct the perpendicular bisector of \(YZ\).
D. Construct the perpendicular bisector of \(XY\).