Coordinate Geometry - Straight Line

Find the value of $k$ in order that the equation $\;$ $12x^2 + kxy + 2y^2 + 11x - 5y + 2 = 0$ $\;$ may represent a pair of straight lines.

Given equation is: $\;$ $12x^2 + kxy + 2y^2 + 11x - 5y + 2 = 0$ $\;\;\; \cdots \; (1)$

Comparing equation $(1)$ with the general second degree equation

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ $\;\;\; \cdots \; (2)$ $\;$ gives

$a = 12, \; h = \dfrac{k}{2}, \; b = 2, \; g = \dfrac{11}{2}, \; f = \dfrac{-5}{2}, \; c = 2$

Equation $(2)$ represents a pair of straight lines if

$abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ $\;\;\; \cdots \; (3)$

Substituting the values of $\;$ $a, \; b, \; c, \; f, \; g, \; h \;$ in equation $(3)$ gives

$12 \times 2 \times 2 + 2 \times \left(\dfrac{-5}{2}\right) \times \dfrac{11}{2} \times \dfrac{k}{2} - 12 \times \left(\dfrac{-5}{2}\right)^2 - 2 \times \left(\dfrac{11}{2}\right)^2 - 2 \times \left(\dfrac{k}{2}\right)^2 = 0$

i.e. $\;$ $48 - \dfrac{55 k}{4} - 75 - \dfrac{121}{2} - \dfrac{k^2}{2} = 0$

i.e. $\;$ $\dfrac{-175}{2} - \dfrac{55k}{4} - \dfrac{k^2}{2} = 0$

i.e. $\;$ $2k^2 + 55k + 350 = 0$

$\implies$ $k = -10$ $\;$ or $\;$ $k = -17.5$