Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 10 - Section 10.3 - Preparing to Do Analytic Proofs - Exercises - Page 473: 13

Answer

M(0,0) N(r,0) P(s+r,t)

Work Step by Step

First, we find the coordinates of M. From the picture, we see vertex M is located at the origin. Hence the coordinates of M are $(0,0)$. Next, we find the coordinates of N. We are given the $x$-coordinate of N, which is $r$, and from the picture, we see N lies on the $x$-axis, which means the $y$-coordinate of N is $0$. Thus vertex N has coordinates $(r,0).$ Lastly, we find vertex P. We are given that figure MNPQ is a parallelogram, and we know that opposite sides of a parallelogram are parallel and are equal in length. So $\overline{QP}$ is parallel to $\overline{MN}$ and QP=MN. To find the $y$-coordinate of P, we note that $\overline{MN}$ lies along the $x$-axis, so $\overline{MN}$ is a horizontal line. Thus since $overline{QP}$ is parallel to $\overline{MN}$, $\overline{QP}$ is a horizontal line as well. This means that points Q and P must have the same $y$-coordinates. We are given the coordinates of Q are $(s,t)$, so the $y$-coordinate of P is t. To find the $x$-coordinate of P, we use the fact that QP=MN. Now, the length MN of $\overline{MN}$ is the distance between the points M and N. Since M and N lie on the same horizontal line, the distance between them is, simply, the positive difference between their $x$-coordinates, which is r-0=r. Now since QP=MN=r and since Q and P lie on the same horizontal line, the distance between Q and P must equal r. Thus the $x$-coordinate of P must be r units more than the $x$-coordinate of Q. Thus the $x$-coordinate of P is s+r, which means the coodinates of P are $(s+r,t)$.
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