Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Test - Page 138: 18

Answer

$(a)$ See the image below. $(b)$ $d=\sqrt{89}$ $(c)$ $M_{PQ}(1, 3.5)$ $(d)$ $m=\frac{5}{8}$ $(e)$ $y=-1.6x+5.1$ $(f)$ $(x – 1)^2 + (y – 3.5)^2 = 22.25$

Work Step by Step

$(a)$ See the image above. $(b)$ We can find the distance using distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ $d=\sqrt{(5+3)^2+(6-1)^2}=\sqrt{64+25}=\sqrt{89}$ $d=\sqrt{89}$ $(c)$ We can locate the midpoint using midpoint formula: $M_{PQ}=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ $M_{PQ}=(\frac{-3+5}{2}, \frac{1+6}{2})=(1, 3.5)$ $(d)$ Slope $m$ of a line is calculated by dividing change in $y$ by change in $x$ for any given two points of the line. $m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-1}{5+3}=\frac{5}{8}$ $m=\frac{5}{8}$ $(e)$ Using the perpendicular bisector form of a line $(x-x_1)^2+(y-y_1)^2=(x-x_2)^2+(y-y_2)^2$ We can write: $(x+3)^2+(y-1)^2=(x-5)^2+(y-6)^2$ $x^2+6x+9+y^2-2y+1=x^2-10x+25+y^2-12y+36$ $10y=-16x+51$ $y=-1.6x+5.1$ $(f)$ To find the equation of the circle, we will use the following form of a circle: $(x – a)^2 + (y – b)^2 = r^2$ Where $r$ is a radius of the circle and $a$ and $b$ are coordinates of the center of the circle; in our case, the coordinates of $M$. We also know the diameter of the circle (The distance between $P$ and $Q$), so the radius is its half. $r=\frac{\sqrt{89}}{2}$ $(x – 1)^2 + (y – 3.5)^2 = (\frac{\sqrt{89}}{2})^2$ $(x – 1)^2 + (y – 3.5)^2 = \frac{\sqrt{89}}{4}$ $(x – 1)^2 + (y – 3.5)^2 = 22.25$
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