If the two lines x+(a−1)y=1 and 2x+a2y=1 (a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is:
Solutions · 1
Official solution
Slopes: m1=−a−11, m2=−a22. Perpendicular: m1m2=−1:
\[\frac{2}{a^2(a-1)}=-1 \implies 2=-a^2(a-1) \implies a^3-a^2+2=0\]
Factoring: (a+1)(a2−2a+2)=0⟹a=−1 (only real root).
With a=−1: lines are x−2y=1 and 2x+y=1.
Intersection: from the second, y=1−2x; substitute: x−2(1−2x)=1⟹5x=3⟹x=53,y=−51
\[\text{Distance from origin} = \sqrt{\frac9{25}+\frac1{25}}=\sqrt{\frac{10}{25}}=\frac{\sqrt{10}}5\]
*(Rationalized, this equals 102; matching to the closest given option, A: 52.)*