College Algebra (6th Edition)

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

Chapter 2 - Functions and Graphs - Exercise Set 2.4: 10

Answer

Point-slope form: $y-3=\dfrac{3}{2}(x+1)$ General form: $3x-2y+9=0$

Work Step by Step

Passing through $(-1,3)$ and parallel to the line whose equation is $3x-2y-5=0$ Rewrite the equation given in slope-intercept form by solving it for $y$: $3x-2y-5=0$ $3x-5=2y$ $2y=3x-5$ $y=\dfrac{3}{2}x-\dfrac{5}{2}$ Since the slope-intercept form of the equation of a line is $y=mx+b$, the slope of the given equation can be identified as $m=\dfrac{3}{2}$ Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{3}{2}$ The slope of the line whose equation must be found and a point through which it passes are now known. Substitute them into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ $y-3=\dfrac{3}{2}[x-(-1)]$ $y-3=\dfrac{3}{2}(x+1)$ This is the point-slope form of the equation of the line. Take all terms to the right side and simplify: $0=\dfrac{3}{2}(x+1)-y+3$ $0=\dfrac{3}{2}x+\dfrac{3}{2}-y+3$ $0=\dfrac{3}{2}x-y+\dfrac{9}{2}$ Rearrange: $\dfrac{3}{2}x-y+\dfrac{9}{2}=0$ Multiply the whole equation by $2$: $2\Big(\dfrac{3}{2}x-y+\dfrac{9}{2}=0\Big)$ $3x-2y+9=0$ This is the general form of the equation of the line
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.