Answer
$y=-\dfrac{1}{2}x - \dfrac{11}{2}$
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) 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:
$y=-\dfrac{1}{2}x - \dfrac{11}{2}$
