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find the equation of a straight line parallel to 2x+3y+11=0 and which is such that sum of its intercepts on the axes is 15

Given, a line l is parallel to a straight line  m whose equation is 2x+3y+11=0. 
Since the line l is parallel to the the line m, its equation will be 2x+3y+C=0.
In slope intercept form, the equation of the line l would be x/a + y/b = 1 ---- (2).
Given a+b=15 or a=15-b.
Therefore , the equation (2) can be written as 
x/(15-b) + y/b = 1
=>  bx + (15-b)y = (15-b)b
=>  bx + (15-b)y - (15-b)b = 0
Comparing the above with eq. (1), we get b = 2k, and (15-b)=3k for some factor k.
Solving for b, we get b = 6 and therefore a = 15-6 = 9. 
Thus, the equation of the line l is x/9 + y/6 =1 or 2x +3y -18 =0

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how did you solve and get b=6?
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Please explain in detail
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What is k?
 
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Factor
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From where you are solving for b
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Find the slope of the given equation, put in the slope/intercept form:
2x + 3y = 11
3y = -2x + 11
y = x + 
parallel lines have the same slope, find the parallel line
Given that intercepts: x + y = 15, therefore
y = (15-x)
the y intercept occurs when x = 0; therefore we can say
y = x + (15-x) = 0
x = -(15-x)
multiply both sides by -1
x = 15 - x
add x to both sides
x = 15
x = 15 *
x = 9 is the x intercept
and
15-9 = 6 is the y intercept
:
See what that looks like

Green line is our parallel line; intercepts x=9; y=6 
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this is the solution. I have made the steps quite simple so that you guys can understand easily. you are welcome!!

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find the equation of the sides of an equilateral triangle whose vertex is (-1,2) and base is y=0
 
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