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Answer 1 · 2016-11-30T08:31:21

#y=5/2x-19/2#

Or #5x -2y = 19#

Conditions:
• Parallel to #5x-2y=2#
• Passes through the point #(3;-2)#

General equation for a line: #y=mx+n# where #m# is the slope of the line.

Rearranging the equation; #2y=5x-2 ; " "y= 5/2x-1#

Here #m= 5/2#, the slope of the required line
(since new line is PARALLEL the slope is the same)

#y=5/2x+n# where n is a numerical constant.

To find #n#, we know the line passes through the given point.

Replacing the values of this point in the equation of the new line:

#-2=(5(3))/2 +n " " rarr n=-2-15/2=-19/2#

#rarr y=5/2x-19/2" "# this is slope-intercept form

Multiplying by 2 to eliminate the denominator;

#2y=5x-19 " or " 5x -2y = 19" "# this is standard form.