Question 188946
Find the equation, in standard form, of the line perpendicular to 2x - 3y = -5 and passing through (3, -2). Write the equation in standard form, with all integer coefficients. 
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1st, find the slope of 2x - 3y = -5.  To find the slope, m, put the eqn in the slope-intercept form, y = mx + b.  All this means is solve for y.
2x - 3y = -5
-3y = -2x - 5
y = (2/3)x + 5/3
The slope is 2/3
The slope of lines perpendicular will be the negative inverse, or -3/2.
Then use:
y-y1 = m*(x-x1) where (x1,y1) is the point (3,-2)
y+2 = (-3/2)*(x-3)
2y + 4 = -3x + 9
3x + 2y = 5 
That's all.