How do you prove that a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects? will ACTUALLY GIVE YOU AND EVERYTHING​

Step-by-step explanation:

If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP is the perpendicular bisector of AB, then CA = CB. If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Consider the line segment AB,

let M be the midpoint of AB, so AM=MA

erect the perpendicular MT to AB from point M. Pick a point P on MT and join it to the points A and B

The triangles PMA and PMB are congruent from the Side Angle Side congruence postulate:

AM=MA, PM is common and m(PMA)=m(PMB)=90°, as MT is perpendicular to AB

so PA=PB

Explanation: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If CP is the perpendicular bisector of AB, then CA = CB. If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Brailiest is appreciated

answer: c

(2, -4)

goes 7 to the

(9, 4)

easy

thus, the first answer choice, y = sec x, is the correct one