Linear algebra 3 lines in r2 Unique solution

[madah12]

Homework Statement

ax+by=k
cx+dy=l
ex+fy=m

If in Exercise 12 k=l=m=0, explain why the system must be consistent. What can be said about
the point of intersection of the three lines if the system has exactly one solution?

Homework Equations

The Attempt at a Solution

ofcourse the system is consisten because x,y=0 is always a solution
but for the second part
all i did was try to get it to reduced row echelon form and what i got is that in the last colum and last row
of augmented matrix m-(f(l-kc/a)/(d-bc/a))
so i said f=0 , l- kc/a = 0 , m-(f(l-kc/a)/(d-bc/a)) = 0 for there to be unique solution i don't know what that would mean about the point but most likely i did mistake in reduced row echelon because too many terms is there any other way to do this?

i know we must have a 0 because if not then 0x +0y = number which can't be

 

[madah12]

oh and i got x= k- b(l-kc/a)/d-bc/a
y= (l-kc/a )/ (d - bc/a)

 

[madah12]

oh one thing i see is that l-kc/a = 0 , l/k = c/a , which means there must be proportionality between the x coefficient and the y intercept

 

[HallsofIvy]

The line ax+ by+ cz= 0 passes through the point (0, 0, 0) for any a, b, c. So if the system is solvable for one point of intersection, that point must be ?

 

[madah12]

The origin? but ax+ by+ cz= 0 is a plane i thought? and we are in r2 not r3 which makes me more confused
but we have this
ax+by=k
cx+dy=l
ex+fy=m
which is different?

 

[madah12]

wait wait so are we still considering the first part where k=l=m=0 in the second part? cause i mean ofcourse we will get x=y=0 if that's the case

 

[SammyS]

As you said, x = y = 0 is a solution to the system. If the system has exactly one solution, then x = y = 0 must be the only solution.

 

[madah12]

yea sorry i didnt know that we were still considering k=l=m=0 that's why i got confused