College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.3 - Lines - 2.3 Assess Your Understanding: 52

Answer

$\color{blue}{y=\dfrac{1}{5}x+\dfrac{23}{5}}$.

Work Step by Step

RECALL: (1) The slope-intercept form of a line's equation is: $y=mx+b$ where $m$=slope and $b$ = y-intercept. (2) The slope of a line that contains the points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the formula: $m=\dfrac{y_2-y_1}{x_2-x_1}$ Solve for the slope of the line using the two given points on the line to obtain: $m=\dfrac{4-5}{-3-2} \\m=\dfrac{-1}{-5} \\m=\dfrac{1}{5}$ Thus, the tentative equation of the line is: $y=\dfrac{1}{5}x+b$ To find the value of $b$, substitute the x and y values of the point $(2, 5)$ into the tentative equation above to obtain: $y=\dfrac{1}{5}x+b \\5=\dfrac{1}{5}(2)+b \\5=\dfrac{2}{5}+b \\5-\dfrac{2}{5}=b \\\dfrac{25}{5}-\dfrac{2}{5}=b \\\dfrac{23}{5}=b$ Therefore, the equation of the line is $\color{blue}{y=\dfrac{1}{5}x+\dfrac{23}{5}}$.
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