Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 1.1 The Distance and Midpoint Formulas - 1.1 Assess Your Understanding - Page 7: 50

Answer

$\left( 0,\,-3+2\sqrt{5} \right)$ and $\left( 0,\,-3-2\sqrt{5} \right)$

Work Step by Step

The points on the $y-$axis are represented by the point$\left( 0,y \right)$. By using the distance formula with \[{{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,-3 \right)\] , ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,y \right)$ and \[d\left( {{P}_{1}},\,{{P}_{2}} \right)=6\], $6=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}}$ By simplifying, $6=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}}$ Solving for $y$, $6=\sqrt{16+{{\left( y+3 \right)}^{2}}}$ Squaring on both sides, ${{6}^{2}}=16+{{\left( y+3 \right)}^{2}}$ $36=16+{{\left( y+3 \right)}^{2}}$ Subtract $16$on both sides, $36-16=16-16+{{\left( y+3 \right)}^{2}}$ $\Rightarrow 20={{\left( y+3 \right)}^{2}}$ $\Rightarrow {{\left( y+3 \right)}^{2}}=20$ By square root method, $\Rightarrow \left( y+3 \right)=\pm \sqrt{20}$ $\Rightarrow \left( y+3 \right)=\pm 2\sqrt{5}$ Subtract $3$ on both sides, $\Rightarrow y=-3\pm 2\sqrt{5}$ $\Rightarrow y=-3+2\sqrt{5}$ and $y=-3-2\sqrt{5}$ Therefore, the points are $\left( 0,\,-3+2\sqrt{5} \right)$ and $\left( 0,\,-3-2\sqrt{5} \right)$.
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