If the equation `2x^2+kxy-3y^2-x-4y-1=0` represents a pair of lines then the value of 'k' can be:

Explanation

For a pair of straight lines, we have,

`Delta = abc+2fgh-af^2-bg^2-ch^2 = 0`

In the given equation,

`a = 2, b = -3,c = -1,g = -1/2, f = -2, h = k/2`

So, putting these values in the above equation,

`(2)(-3)(-1)+2(-2)(-1/2)(k/2)-(2)(-2)^2-(-3)(-1/2)^2-(-1)(k/2)^2 = 0`
`=>6+k-8+3/4+k^2/4 = 0`

`=>k^2+4k -5 = 0`

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