College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.5 - Page 653: 54

Answer

$y=\displaystyle \frac{1}{3}x+\frac{10}{3}$

Work Step by Step

$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{array}\right|$ = see p.645... $=a_{1}b_{2}c_{3}+b_{1}c_{2}a_{3}+c_{1}a_{2}b_{3}-a_{3}b_{2}c_{1}-b_{3}c_{2}a_{1}-c_{3}a_{2}b_{1}$ --------------------- Using the given formula with $(x_{1},y_{1})=(-1,3)$ $(x_{2},y_{2})=(2,4)$ $\left|\begin{array}{lll} x & y & 1\\ -1 & 3 & 1\\ 2 & 4 & 1 \end{array}\right|=0$ $3x+2y+(-4)-6-4x-(-y)=0$ $-x+3y-10=0$ For slope-intercept form, this equation is solved for y: $3y=x+10\qquad/\div(3)$ $y=\displaystyle \frac{1}{3}x+\frac{10}{3}$
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