## College Algebra 7th Edition

$y=-\dfrac{1}{2}x - \dfrac{11}{2}$
RECALL: (1) The slope-intercept form of a line's equation is $y=mx+b$ where m = slope and b = y-intercept. (2) Parallel lines have equal slopes. Transform $x+2y=6$ to slope-intercept form to obtain: $x+2y=6 \\2y=-x+6 \\\dfrac{2y}{2}=\dfrac{-x+6}{2} \\y = -\dfrac{1}{2}x +3$ The slope of this line is $-\dfrac{1}{2}$. The line we are looking for the equation of is parallel to the line above. Thus, the slope of the line is also $-\dfrac{1}{2}$. This means that a tentative equation of the line is: $y=-\dfrac{1}{2}x+b$ The line passes through the point $(1, -6)$. To find the value of $b$, substitute the x and y coordinates of this point into the tentative equation above to obtain: $y=-\dfrac{1}{2}x+b \\-6 = -\dfrac{1}{2}(1) + b \\-6 = -\dfrac{1}{2} + b \\-6 + \dfrac{1}{2} = b \\-\dfrac{12}{2} + \dfrac{1}{2} = b \\-\dfrac{11}{2} = b$ Therefore, the equation of the line we are looking for is: $y=-\dfrac{1}{2}x - \dfrac{11}{2}$