Finding the angle between the lines represented by $9x^2+24xy+16y^2=0$ 
Question: The angle between the lines represented by 
  $$9x^2+24xy+16y^2=0$$
  is
(A) $90^\circ$
(B) $0^\circ$
(C) $180^\circ$
(D) None of Above

I’ve tried to solve it by making L.H.S a perfect square and then taking the square root to find the slope in order to get an angle. But the answer i’m getting is not correct. 
Could anybody help me here? Thanks in advance.
 A: Observe that $$...\iff (3x+4y)^2=0\iff 4y=-3x$$
Which is a line. Can you finish now?
A: We know that the equation $$ax^2+2hxy+by^2=0$$ represents a pair of straight lines passing through origin and angle btw them is given by 
$$Tan θ = [2√(h^2-ab)]/(a+b) $$
$$Cos θ = (a+b)/√[(a-b)^2 + 4h^2]$$........  using $cosθ$ to avoid complication if $θ$  is  $90$
so here from your problem.....$$a=9::b=16::h=12$$
putiing respective values in $cosθ$ .....we get $$cosθ = 1$$
Thus $$θ=0$$ .......so $$option    {B}$$
       Now you may ask how to derive the formulae for angle btw pair of straight lines

      The equation ax2+2hxy+by2=0 ,
      can be written as product of two linear factors, ax2+2hxy+by2= (lx+my)(px+qy)

      breaking the following we get, lp=a, mq=b and lq+mp=2h
      and the separate equations of lines are lx+my=0 and px+qy=0
      therefore the angle btw the lines...

      Cos θ = (lp+mq)/√[(l2+m2)(p2+q2)]
      = (lp+mq)/√[(lp-mq)2 +(lq+mp)2]
      = (a+b)/√[(a-b)2 + 4h2]




      P.S: sorry for not writing the the solution in latex format

