Question

The point of concurrent of the lines $(3k+1)x-(2k+3)y+(9-k)=0$

• A. $(1,1)$
• B. $(1,-1)$
• C. $(3,4)$
• D. None of these

Answer 1

The correct option is C $(3,4)$

$(3k+1)x-(2k+3)y+9-k=0$

$⟹(x-3y+9)+k(3x-2y-1)=0$

This equality holds for every $k$ then we have,

$x-3y+9=0⟹x-3y=-9$

$3x-2y-1=0⟹3x-2y=1$

Solving above two equations we get,

$y=4$ and $x=3y-9=3\left(4\right)-9=3$

So, $(x,y)=(3,4)$.