**General Equation of Second Degree**

The necessary and sufficient condition for ax² + 2hxy + by² + 2gx + 2fy + c = 0 to represent a pair of straight lines is that abc + 2fgh – af² – bg² – ch² = 0 or \(\left| \begin{matrix} a & h & g \\ h & b & f \\ g & f & c \\\end{matrix} \right|=0\).

**Equation of the Bisectors: **The equations of the bisectors of the angles between the lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 are given by \(\frac{{{\left( x-{{x}_{1}} \right)}^{2}}-{{\left( y-{{y}_{1}} \right)}^{2}}}{a-b}=\frac{\left( x-{{x}_{1}} \right)\left( y-{{y}_{1}} \right)}{h}\), where (x₁, y₁) is the point of intersection of the lines represented by the given equation.

**Problem: **For what value of λ does the equation 6x² – 42xy + 60y² – 11x + 10y + λ = 0 represent two straight lines?

**Solution: **Comparing the given equation with ax² + 2hxy + by² + 2gx + 2fy + c = 0 we obtain a = 6, b = 60, c = λ, h = -21, g = -11/2.

The given equation will represent a pair of straight lines if abc + 2fgh – af² – bg² – ch² = 0,

6(60) + λ + 2 (5) (-11/2) (-21) – 6 (5)² = 0,

λ = -10.