If `2x + 3y = 7` makes equal angles with pair of lines `9x^2 +12xy + ky^2=0` then `k=` (i)`7` (ii)`14` (iii)`-14` (iv)`-7`

The angle between the pair of lines represented by 2x^2-7xy+3y^2=0 is

If the lines 2 x^(2)+7 x y+3 y^(2)-9 x-7 y+k=0 represents a pair of lines, then k=

If 2x^(2)+xy-3y^(2)+4x+ky-6=0 represents a pair of lines then k=

Equation of pair of lines passing through origin and making an angle (pi)/(4) with the line 3x+2y+4=0 is (3x+2y)^(2)-k(2x-3y)^(2)=0 then k=(i)1 (ii) -1 (iii) 3 (iv) (1)/(3)

The figure formed by the pairs of lines 12x^(2)+7xy-12y^(2)=0 and 12x^(2)+7xy-12y^(2)+10x+5y+2=0 is a

If the equation 2x^(2) + 7xy+3y^(2) - 9x - 7y+k=0 represents a pair of lines, then kis equal to

Solve 3x + 2y = 14 and - x + 4y = 7 and hence find the value of k for which 3x = 2ky + 6.