The vertices of a triangle `OBC` are `O(0,0),B(-3,-1),C(-1,-3)`. Equation of line parallel to `BC` & intersecting the sides `OB` & `OC` whose perpendicular distance from the point `(0,0)` is `1/(sqrt(2))` is `ax+by+2=0` then the value of `(a^(4)+b^(4))/4` is

Answer

8

Explanation

Slope of `BC=-1` slope of line parallel to `BC=-a/b`

`:.=(-a)/b =-1`
`impliesa=b`
Distance of line `ax+by+2=0` from the origin
`=2/(sqrt(a^(2)+b^(2)))=1/(sqrt(2))impliesa^(2)+b^(2)=8`
`:.a^(2)=b^(2)=4`
`:.a=b=2`