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<andλ>2 ∴
λ∊(2,)....(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]