How do you write an equation in standard form for a line passing through (3, 4) and is parallel to the line y = -x + 6?

1 Answer
May 2, 2015

Answer: #x + y = 7#

Explanation:

The standard form equation for a line on the #xy#-plane is #Ax + By = C#, where #A, B, C# are constants. However, this problem may be easier if we begin in slope-intercept form (i.e. #y = mx+b#), since the equation of the parallel line is given in said form.

For two lines to be parallel in the #xy#-plane, they must have the same slope (represented by #m# in slope-intercept form) and different Y-intercepts. The line parallel to our desired line is given the equation #y = -x +6#, which could be rewritten as # y = (-1)x + 6# due to the identity property of multiplication. Our parallel line thus has a slope of #m = -1#, meaning that our desired line possesses the same slope.

At this point, we may construct our desired line in either slope-intercept or (by skipping a step) standard form. We will begin with slope-intercept and then convert to standard form. Our desired line, given our discovery above, has an equation in slope-intercept form of #y = -x + b#, where #b# is an unknown Y-intercept. We are given the coordinates for a point on the line #(3,4)#, and may substitute these coordinates in for #x# and #y# to determine #b#:
#y = -x + b -> 4 = -3 + b -> 7 = b#

Having determined the Y-intercept, we now know that our line has a slope-intercept equation of #y = -x + 7#. By adding #x# to either side-

#y = -x + 7 -> x + y = 7#

-we obtain the standard form equation of our line.