Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - 9-6 The Quadratic Formula and the Discriminant - Practice and Problem-Solving Exercises - Page 588: 54

Answer

$n=-1$.

Work Step by Step

The given points are $(x_1,y_1)=(n,6)$ and $(x_2,y_2)=(1,2)$ Slope $m_1=\frac{y_2-y_1}{x_2-x_1}$. Substitute all values. Slope $m_1=\frac{2-6}{1-n}=\frac{-4}{1-n}$. The given equation of the parallel line: $2x+y=3$ The slope intercept form is $y=-2x+3$. Slope $m_2=-2$. Equate both slopes. $\Rightarrow m_2=m_1$ Substitute both values. $\Rightarrow -2=\frac{2-6}{1-n}$ Multiply the equation by $(1-n)$. $\Rightarrow -2\cdot (1-n)=\frac{-4}{1-n}\cdot (1-n)$ Simplify. $\Rightarrow -2+2n=-4$ Add $2$ to both sides. $\Rightarrow -2+2n+2=-4+2$ Simplify. $\Rightarrow 2n=-2$ Divide both sides by $2$. $\Rightarrow \frac{2n}{2}=\frac{-2}{2}$ Simplify. $\Rightarrow n=-1$.
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