Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.3 - Lines, Circles, and Parabolas - Exercises A.3 - Page AP-17: 10


$y=-\displaystyle \frac{1}{5}x+\frac{23}{5}$

Work Step by Step

$A=(x_{1},y_{1})=(3, \ 4)$ $B=(x_{2},y_{2})=(-2, \ 5)$ The increments in the coordinates are calculated as $\left\{\begin{array}{l} \Delta x=x_{2}-x_{1}\\ \Delta y=y_{2}-y_{1} \end{array}\right.$ $\left\{\begin{array}{l} \Delta x=-2-3=-5\\ \Delta y=6-4=1 \end{array}\right.$ Neither of the increments is zero; so the line is neither vertical nor horizontal. The slope of the line that passes through A and B, if the line is nonvertical, ($\Delta x\neq 0)$ is calculated as $m=\displaystyle \frac{\Delta y}{\Delta x}=\frac{1}{5}==-\frac{1}{5}$, The point-slope equation of a line containing the point $(x_{1},y_{1})$, with slope $m$ is$\quad y=y_{1}+m(x-x_{1})$ Given $(x_{1},y_{1})=(3,4)$ and $m=-\displaystyle \frac{1}{5}$ $y=4+(-\displaystyle \frac{1}{5})(x-3)$ $y=4-\displaystyle \frac{1}{5}x+\frac{3}{5}$ $y=-\displaystyle \frac{1}{5}x+\frac{20+3}{5}$ $y=-\displaystyle \frac{1}{5}x+\frac{23}{5}$
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