Question 709407
p must be chosen just so that the slopes of the two lines are different.  That condition is p<-10 and p>-10.  p can be any real number, just not -10.


Your equations:
7x-5y-4=0 and 14x+py+4=0


The first one, -5y=-7x+4, y=(7/5)x+4, the slope is (7/5).

The second equation, py=-14x-4, y=(-14/p)x-4, the slope is variably (-14/p).

If the slopes of the two lines are DIFFERNT, then the system will have ONE solution.  If p=-10, then the second line has slope (-14/-10)=7/5, and is parallel to the first line and then the system has no solution.  


The only requirement for the two lines meeting in one point is that p NOT equal -10.