Question 28830
write equation of line described.
through (-8,-16), parallel to 7x-5y=-16

The given line is 7x-5y=-16
That is 7x-5y +16 =0  ----(1)
Any line parallel to (1) will be of the form 
    7x-5y +k =0  ----(2)
[every value of k will produce a line in the system (2) which is parallel to (1)
Our line is parallel to (1) and passes through the fixed point P(-8,-16)
This means that line in the system (2) which passes through the fixed point 
P(-8,-16)will yield a particular k which when supplied in (2) will give our required line.
Therefore putting x=-8 and y=-16 in (2)
7X(-8)-5X(-16) + k = 0
-56+80+k=0
24+k=0
k=-24  ----(*)
Using (*) in (2),that is putting k=-24 in (2),
we get the required line given by
7x-5y-24=0
Verification: P(-8,-16) should be a point on this line
Therefore LHS = 7x-5y-24 = 7X(-8) - 5X(-16) -24 = -56+80-24 =80-80= 0 =RHS
Note:[given a line Ax+By+C=0 any line parallel to this is of the form Ax+By +D=0
Why are the x and y parts the same?
It is because when lines are parallel slopes are equal and the equal slope here is (-A/B)]