Question 189027
find the equation of the line that contains the point (-1, -3) and is perpendicular to the graph of 3x-5y=2
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Step 1, find the slope, m, of the line.
To do that, put the eqn in the slope-intercept form, which means solve for y.
3x-5y=2 --> y = (3/5)x - 2/5
In this form, it's y = mx + b, where m is the slope and b in the y-intercept.
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Lines parallel to this  line have the same slope.  The slope of lines perpendicular have a slope that's the negative inverse, m = -5/3
Step 2:
Use y-y1 = m*(x-x1) where (x1,y1) is the point (-1,-3)
y+3 = (-5/3)*(x+1)
y = (-5/3)x - 14/3 (slope-intercept form)
5x + 3y = -14 (standard form)