If P1=(x1, y1), P2=(x2, y2) and M=((x1+x2)/(2), (y1+y2)/(2)), show that d(P1, M)=d(M, P2)=(1)/(2

Name: Problem 61, Chapter 2: Precalculus
Uploaded: 2021-07-30T14:18:29-08:00
Duration: 4 min
Channel: Yujie Wang
Description: If P1=(x1, y1), P2=(x2, y2) and M=((x1+x2)/(2), (y1+y2)/(2)), show that d(P1, M)=d(M, P2)=(1)/(2) d(P1, P2) . (This is one step in the proof of Theorem 2.)

step 1 Okay, so let's look at this problem. Okay, so we have three points, P1, P2, and M, okay? So now

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Question

If $P_{1}=\left(x_{1}, y_{1}\right), P_{2}=\left(x_{2}, y_{2}\right)$ and $M=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$, show that $d\left(P_{1}, M\right)=d\left(M, P_{2}\right)=\frac{1}{2} d\left(P_{1}, P_{2}\right) .$ (This is one step in the proof of Theorem 2.)

   If $P_{1}=\left(x_{1}, y_{1}\right), P_{2}=\left(x_{2}, y_{2}\right)$ and $M=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$,
show that $d\left(P_{1}, M\right)=d\left(M, P_{2}\right)=\frac{1}{2} d\left(P_{1}, P_{2}\right) .$ (This is one step in the proof of Theorem 2.)

Precalculus

Raymond A. Barnett,… 7th Edition

Chapter 2, Problem 61

Instant Answer

Solved by Expert Yujie Wang

07/30/2021

Step 1

The distance formula is given by $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$. Substituting the coordinates of $P_{1}$ and $M$ into the formula, we get: \[d\left(P_{1}, Show more…

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If $P_{1}=\left(x_{1}, y_{1}\right), P_{2}=\left(x_{2}, y_{2}\right)$ and $M=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$, show that $d\left(P_{1}, M\right)=d\left(M, P_{2}\right)=\frac{1}{2} d\left(P_{1}, P_{2}\right) .$ (This is one step in the proof of Theorem 2.)

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Key Concepts

Midpoint

The midpoint of a line segment in coordinate geometry is defined as the point whose coordinates are the averages of the corresponding coordinates of the endpoints. It represents the exact center of the segment, ensuring that the segment is divided into two congruent parts.

Distance Formula

The distance formula computes the distance between two points in the Euclidean plane by taking the square root of the sum of the squared differences in their coordinates. This formula, which is based on the Pythagorean theorem, is essential for quantifying the length of segments in coordinate geometry.

Equidistance Property of the Midpoint

A fundamental property of the midpoint is that it is equidistant from the endpoints of the line segment. This characteristic implies that the distances from the midpoint to each endpoint are equal, which in turn leads to the conclusion that each distance is exactly half of the total distance between the endpoints.

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