Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - Summary Exercises on Graphs, Circles, Functions, and Equations - Exercises - Page 247: 3

Answer

$\color{blue}{\bf\text{(a) }5}$ $\color{blue}{\bf\text{(b) }(\frac{1}{2},2)}$ $\color{blue}{\bf\text{(c) }y=2}$

Work Step by Step

$\bf{(a)}$ distance between points $\bf{P(-2,2)}$ and $\bf{Q(3,2)}$ Use the distance formula: $$d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}$$ $d=\sqrt{(-2-3)^2+(2-2)^2}$ $d=\sqrt{5^2+0^2}$ $d=\sqrt{5^2}$ $d=\color{blue}{\bf{5}}$ $\bf{(b)}$ midpoint between points $\bf{P(-2,2)}$ and $\bf{Q(3,2)}$ Use the midpoint formula: $$m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$$ $m=(\frac{-2+3}{2},\frac{2+2}{2})$ $m=(\frac{1}{2},\frac{4}{2})$ $m=\color{blue}{\bf(\frac{1}{2},2)}$ $\bf{(c)}$ equation for the line that passes through the points $\bf{P(-2,2)}$ and $\bf{Q(3,2)}$ in slope intercept form Use point slope form, $y−y_1=m(x−x_1)$ where $\bf{m}$ = slope = $\frac{\Delta{y}}{\Delta{x}} =\frac{y_1−y_2}{x_1−x_2}$ $$y−y_1=\frac{y_1−y_2}{x_1−x_2}(x−x_1)$$ $y−2=\frac{2-2}{-2-3}(x−(-2))$ $y−2=\frac{0}{-5}(x+2)$ $y−2=0(x+2)$ $y−2=0$ $\color{blue}{\bf{y=2}}$ $\color{blue}{\bf\text{(a) }5}$ $\color{blue}{\bf\text{(b) }(\frac{1}{2},2)}$ $\color{blue}{\bf\text{(c) }y=2}$
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