Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.7 - The Coordinate Plane - Exercises - Page 38: 18

Answer

$k $ =$ \frac{1}{2}$ OR $0.5$

Work Step by Step

Distance 'd' between two points $(x_{1},y_{1}) $ and $(x_{2},y_{2})$ is given by- d = $\sqrt{(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2}} $ Now, Distance '$d_{1}$' between points $(k,k) $and $(-1,0)$ will be - $d_{1}$ = $\sqrt{(k - (-1))^{2} + (k - 0)^{2}} $ i.e. $d_{1}$ = $\sqrt{(k + 1)^{2} + k^{2}} $ i.e. $d_{1}$ = $\sqrt{k^{2} +1 +2k + k^{2} } $ i.e. $d_{1}$ = $\sqrt{2k^{2} +2k +1 } $ Now, Distance '$d_{2}$' between points $(k,k) $and $(0,2)$ will be - $d_{2}$ = $\sqrt{(k - 0)^{2} + (k - 2)^{2}} $ i.e. $d_{2}$ = $\sqrt{k^{2} + k^{2} - 4k +4 } $ i.e. $d_{2}$ = $\sqrt{2k^{2} - 4k +4 } $ Now According to problem- $d_{1}$ = $d_{2}$ i.e. $\sqrt{2k^{2} +2k +1 } $ = $\sqrt{2k^{2} - 4k +4 } $ Squaring on both sides- $2k^{2} +2k +1 $ = $2k^{2} - 4k +4 $ i.e. $2k +1 $ = $ - 4k +4 $ (Subtracting $2k^{2}$ from both sides) i.e. $2k +4k $ = $ 4-1 $ i.e. $6k $ = $ 3$ i.e. $k $ = $ \frac{3}{6} = \frac{1}{2}$ OR $0.5$ i.e. $k $ =$ \frac{1}{2}$ OR $0.5$
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