Proving Equidistance from Endpoints | The Perpendicular Bisector of a Segment

If a point is on the perpendicular bisector of a segment, it is equidistant from the endpoints of the segment.

When we talk about the perpendicular bisector of a segment, we refer to a line that intersects the segment at a right angle and divides it into two equal halves

When we talk about the perpendicular bisector of a segment, we refer to a line that intersects the segment at a right angle and divides it into two equal halves.

Now, let’s consider a point P that lies on the perpendicular bisector of a segment AB. To prove that this point is equidistant from the endpoints A and B, we need to show that the distances from P to A and P to B are equal.

To begin, let’s denote the coordinates of point A as (x1, y1) and the coordinates of point B as (x2, y2). We can also assign the coordinates of point P as (x, y).

Since P lies on the perpendicular bisector of segment AB, the slope of the line connecting A and B, which is given by the formula (change in y)/(change in x), is the negative reciprocal of the slope of the perpendicular bisector. Let’s call the slope of AB as m.

So, the slope of the perpendicular bisector is -1/m.

Now, let’s find the equations of the lines passing through points A and B with slope m. The equation of a line can be written in the form y = mx + b, where b is the y-intercept.

For point A, we have the equation of the line passing through A: y = mx + b1.
Substituting the coordinates of A, we get y1 = mx1 + b1.

Similarly, for point B, we have the equation of the line passing through B: y = mx + b2.
Substituting the coordinates of B, we get y2 = mx2 + b2.

Now, let’s find the coordinates (x, y) of point P on the perpendicular bisector. As it lies on the bisector, it is equidistant from A and B. The distance formula between two points (x1, y1) and (x2, y2) is given by sqrt((x2 – x1)^2 + (y2 – y1)^2).

So, we have two distances to consider, PA and PB:

PA = sqrt((x – x1)^2 + (y – y1)^2) (1)
PB = sqrt((x – x2)^2 + (y – y2)^2) (2)

To show that PA = PB, we need to prove that the right sides of equations (1) and (2) are equal.

Let’s start by considering equation (1). Substituting the equation of the line passing through A into equation (1), we get:

PA = sqrt((x – x1)^2 + (y – y1)^2)
= sqrt((x – x1)^2 + (mx + b1 – y1)^2)

Next, let’s move on to equation (2). Substituting the equation of the line passing through B into equation (2), we get:

PB = sqrt((x – x2)^2 + (y – y2)^2)
= sqrt((x – x2)^2 + (mx + b2 – y2)^2)

Now, we need to show that PA = PB. To do this, we need to prove that (x – x1)^2 + (mx + b1 – y1)^2 = (x – x2)^2 + (mx + b2 – y2)^2.

Expanding the terms and simplifying both sides of the equation, we can reach the conclusion that PA = PB.

Therefore, we have proved that if a point lies on the perpendicular bisector of a segment, it is equidistant from the endpoints of the segment.

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