College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.1 - The Coordinate Plane - 1.1 Exercises - Page 92: 28

Answer

(a) Refer to the image below for the plot. (b) $d=\sqrt{41}$ (c) $\text{midpoint} = \left(-3.5,-1\right)$
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Work Step by Step

RECALL: (1) The point $(x, y)$ has: $x$ = directed distance of the point from the y-axis (positive when to the right of the y-axis, negative when to the left) $y$ = directed distance of the point from the x-axis (positive when above the x-axis, negative when below) (2) The distance between the points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the distance formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$. (3) The coordinates of the midpoint of the segment joining the points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$. (a) Use the definition in (1) above to plot the given points. (refer to the attached image in the answer part above) (b) Find the distance between the two points using the formula in (2) above to obtain: $d=\sqrt{(-6-(-1))^2+(-3-1)^2} \\d=\sqrt{(-6+1)^2+(-4)^2} \\d=\sqrt{(-5)^2+16} \\d=\sqrt{25+16} \\d=\sqrt{41}$ (c) Find the coordinates of the midpoint of the segment joining the two points using the formula in (3) above to obtain: $\text{midpoint} = \left(\dfrac{-1+(-6)}{2}, \dfrac{1+(-3)}{2}\right) \\\text{midpoint} = \left(\dfrac{-7}{2}, \dfrac{-2}{2}\right) \\\text{midpoint} = \left(-3.5,-1\right)$
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