Question 736809
You ask, "HOW...?"
Good! 


Here is how:
A line perpendicular to 2y-3x+5=0 will have a slope equal negative reciprocal of the slope of {{{2y-3x+5=0}}}.


2y-3x+5=0
{{{2y=3x-5}}}
{{{y=(3/2)x-5/2}}}
Slope of the given line is {{{(3/2)}}}.  The line perpendicular to the given line has a slope of {{{-(2/3)}}}.


The line we want has slope {{{-(2/3)}}} and contains the point A(-5, 9).  Let's use slope-intercept form to find the y-intercept*.
{{{y=-(2/3)x+b}}}
{{{y+(2/3)x=b}}}
{{{b=y+(2/3)x}}}
substitute the coordinate values for A,
{{{b=9+(2/3)(-5)}}}
{{{b=9-10/3}}}
{{{b=27/3-10/3}}}
{{{b=17/3}}}
OR
{{{b=5&2/3}}}


Solution equation to question is {{{highlight(y=-(2/3)x+17/3)}}} OR {{{highlight(y=-(2/3)x+5&2/3)}}}.  


* You can choose point-slope form instead and do appropriate arithmetic steps.  Also, you could use or put your equation into or use from standard form if you wished.