I cannot quite understand the first question. Do you mean point Q is a random point? Q is at (r,s) now, but Q may be at (t,u), or (v,w), or (x,y), or.....?

For the fixed point P,

q = m(p) +c --------------(1)

For the random point Q, which is at (r,s) now,

s = m(r) +c -------------(2)

Eq.(1) minus Eq.(2),

q -s = m(p) -m(r)

q -s = m(p -r)

m = (q-s) /(p -r) ------(i)**

If Q is at another position, say at (x,y),

y = m(x) +c ------------(3)

Eq.(1) minus Eq.(3),

q -y = m(p) -m(x)

q -y = m(p -x)

m = (q -y) /(p -x) ------(ii)**

Now, compare Eq.(i) and Eq.(ii).

The gradient m remains the same even if point Q transferred from (r,s) to (x,y).

Is that what your first question wants to see?

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Your second question.

ax +by +c = 0 --------(4)

Eq.(4) will always be a straight line as long as there is a variable ...x or y... left on the equationa non-xero c is also left in the equation. Meaning,and

ax +c = 0 is a line. ----**

by +c = 0 is a line. ----**

c = 0 is not a line.

ax = 0 is not a line

by = 0 is not a line.

However, ax +by = 0 is a line. ----**

Therefore, if any two of the a,b,c are zeros at the same time, there will be no straight line.

So, a=0, b=0, c = any number -----no straight line.

a = 0, b = any number, c = 0 -------no line.

a = any number, b=0, c=0 ----------no line.