(a) Equation of any line L perpendicular to
5x − y = 1 is x + 5y = k ...(i)
Where k is an arbitrary constant.
If this line cuts x-axis at A and y-axis at B, then for A, y = 0 and from (i) x = k i.e., A is the point (k, 0). For B, x = 0 and from (i) y =
i.e., B is the point
(0,)∴ Area of the given ∆OAB =
=(x1y2−x2y1)=(−0)=But according to the given condition,
=5 or
k2= 50
∴ k =
±5√2Hence, from (i), the required equation of the line is
x + 5y =
5√2or x + 5y =
−5√2