Three lines of triangle are given by (x2−y2)(2x+3y−6)=0 ⇒ (x−y)(x+y)(2x+3y−6)=0 ∴ Three lines of triangle are x−y=0,x+y=0 and 2x+3y−6=0
From given lines of triangle, the required ∆OAB is formed. ∵ (−2,λ) lies inside the triangle. ∴ 2(−2)+3(λ)−6<0‌and−2+λ>0 ⇒−4+3λ−6<0‌and‌λ>2 ⇒3λ<10‌and‌λ>2 ⇒ λ<
10
3
‌and‌λ>2 ∴λ∊(2,
10
3
)....(i) Now, (μ,1) lies outside the triangle. To find value of μ, we find the interval [M, N] for values of x. x+1≥0‌and‌x−1≤0 x≥−1‌and‌x≤1 ∴ x∊[−1,1] ∵ (μ,1) lies outside then triangle. ∴ μ∊(−∞,−1)‌⋃(1,∞) or μ∊R−[−1,1]