If two lines a1x+b1y=c1 and a2x+b2y=c2 are perpendicular, then a1a2+b1b2=0.
Since, given lines x+(a−1)y=1 and 2x+a2y=1 are perpendicular, hence, we get
1⋅2+(a−1)⋅a2=0
⇒a3−a2+2=0
⇒(a+1)⋅(a2−2a+2)=0
⇒a=−1 or a2−2a+2=0
But, the discriminant of the quadratic equation is (−2)2−4×1×2=4−8=−4, hence, no real roots exist.
Thus, the only possible value of a is −1.
And, hence the lines are x−2y=1 and 2x+y=1.
Point of intersection of these two lines is (53,−51)
We know that, the distance of a point (x,y) from origin is x2+y2
Hence, distance of the point (53,−51) from origin =259+251=52.