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: 17

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