How do you find the equation of a line that Contains point (-2, -3) and is perpendicular to 5x-2y=8?

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Answer 1 · 2016-05-28T18:29:50

#y = (-2x)/5 -3 4/5" "# or # y = (-2x)/5 - 19/5#

First change the equation of the given line into standard form,
#y = mx + c#
#2y = 5x - 8 rArr y= ( 5x)/2 - 4#

The slope of this line is #5/2#, so the slope of a line perpendicular to this is #-2/5#

Given a point #(-2, -3)# and the gradient (#-2/5#), the quickest and easiest way to find the equation is to use the formula:

#y - y_1 = m(x - x_1)#

#y - (-3) = (-2)/5(x - (-2))#

#y + 3 = (-2x)/5 - 4/5#

#y = (-2x)/5 -3 4/5#