Question 258389
I had to think a second about this one. 
They are asking for the equation that for the line which passes through (9,5) and is  perpendicular to 5x + 3y =15.
Then we need the intersection of the two lines and then the distance from the intersection to (9,5)

so find the slope of 5x + 3y =15
3y=15-5x
y=15/3-5x/3
m=-5/3
perpendicular slope would be 3/5
5=3/5(9)+b
25=27+5b
-2=5b
-2/5=b
y=3/5x-2/5
5x + 3y =15
5y=3x-2 and 
3x-2=5y

3x-5y=2
5x+3y=15
9x-15y=6
25x+15y=75
34x=81

x = 81/34,   y = 35/34
(9,5)
(81/34,35/34)
a^2+b^2=c^2
sqrt((9-81/34)^2 +(5-35/34)^2)=
7.7 units

Here are some other ways to do it. Some much shorter.
http://www.worsleyschool.net/science/files/linepoint/distance.html