Question

For all real values of a and b lines `(2a + b)x +(a +3b)y + (b-3a) =0 `and `mx+ 2y +6 =0 `are concurrent, then m is equal to (A) -2 (B) -3(C)-4 (D) -5

Answer 1

Correct Answer - -2
Lines (2a+b)x+(a+3b)y+(b-3a) = 0 or a(2x+y-3) +b(x+3y+1)=0 are concurrent at the point of intersection of lines 2x+y-3 = 0 and x+3y+1 =0, which is (2, -1).
Now, line mx+2y+6= 0 mus pass through this point.
Therefore, 2m-2+6=0 or m=-2.