Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 102: 25

Answer

See graph. Distance is 10 and midpoint is at (3, 12)

Work Step by Step

First we plot out our two points as shown in red and connect them. Next we need to make this into a triangle so we draw another point using the x of one point and the y of the other (shown in blue) at (6, 8). To use the pythagorean theorem we need to get the distance between our red points and blue point. So length of A is 16-8 = 8 and length of B = 6-0 = 6 Now we can calculate $A^2 + B^2 = C^2$ $8^2 + 6^2 = C^2$ $64 + 36 = C^2$ $100 = C^2$ $10 = C$ So now we have our distance. Next to get the midpoint we use the formula: $\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}$ Plug in our numbers and we get: $\frac{0+6}{2}, \frac{8+16}{2}$ $\frac{6}{2}, \frac{24}{2}$ Giving us our midpoint coordinates of (3, 12)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.