Given that x+y=|a| and ax−y=1 Case I: If a>0 x+y=a . . . (i) ax−y=1 . . . (ii) On adding equations (i) and (ii), we get x(1+a)=1+a⇒x=1 y=a-1 Since given that intersection point lies in first quadrant So, a−1≥0 ⇒a≥1 ⇒a∈[1,∞) Case II : If a<0 x+y=−a . . . (iii) ax−y=1 . . . (iv) On adding equations (iii) and (iv), we get x(1+a)=1−a x=
1−a
1+a
>0⇒
a−1
a+1
<0 Since a−1<0 ∴a+1>0 ⇒a>−1 . . . (v) ←
0
−1
> y=−a−
1−a
1+a
>0=
−a−a2−1+a
1+a
>0 ⇒−(
a2+1
a+1
)>0⇒
a2+1
a+1
<0 Since a2+1>0 ∴a+1<0 ⇒a<−1 . . . (vi) From (v) and (vi), a∈φ Hence, Case-II is not possible. So, correct answer is a∈[1,∞)