Three lines of triangle are given by (x2−y2)(2x+3y−6)=0 ⇒(x−y)(x+y)(2x+3y−6)=0 ∴ The 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<0 and ‌‌−2+λ>0 ⇒‌‌−4+3λ−6<0 and λ>2 ⇒‌‌3λ<10 and λ>2 ⇒‌‌λ<‌
10
3
and λ>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≥0 and x−1≤0 ⇒‌‌x≥−1 and x≤1 ∴‌‌x∈[−1,1] ∵(µ,1) lies outside the triangle. ∴‌‌µ∈(−∞,−1)∪(1,∞) or ‌‌µ∈R−[−1,1]