Ans.

(i)

Sol. Case 1 : Let the line L cuts AO and AB at distances x and y from A.

⇒ Area of the triangle with sides x and y is

= 12 ⇒ xy = 40

Also, x + y = 12 (using perimeter bisection)

This is not possible.

Case 2 : If the line L cuts OB and BA at distances y and x from B then we have xy = 30 and x + y = 12.

⇒ x = 6 + and y = 6 –

Case 3 : If the line L cuts the sides OA and OB at distance x and y from O then x + y = 12 and xy = 24.

x, y = 6 ± 2 (not possible)

So there is a unique line possible

Let Point P be ( α , β )

Using parametric equation AB

β = 6 – (6 + ) and α = (6 + )

⇒ Slope of PQ is = .

(ii)

Sol. Case 1 : Let the line L cuts AO and AB at distances x and y from A.

⇒ Area of the triangle with sides x and y is

= 12 ⇒ xy = 40

Also, x + y = 12 (using perimeter bisection)

This is not possible.

Case 2 : If the line L cuts OB and BA at distances y and x from B then we have xy = 30 and x + y = 12.

⇒ x = 6 + and y = 6 –

Case 3 : If the line L cuts the sides OA and OB at distance x and y from O then x + y = 12 and

xy = 24.

x, y = 6 ± 2 (not possible)

So there is a unique line possible

Let Point P be ( α , β )

Using parametric equation AB

β = 6 – (6 + ) and α = (6 + )

⇒ Slope of PQ is = .

(iii)

Sol. Case 1 : Let the line L cuts AO and AB at distances x and y from A.

⇒ Area of the triangle with sides x and y is

= 12 ⇒ xy = 40

Also, x + y = 12 (using perimeter bisection)

This is not possible.

Case 2 : If the line L cuts OB and BA at distances y and x from B then we have xy = 30 and x + y = 12.

⇒ x = 6 + and y = 6 –

Case 3 : If the line L cuts the sides OA and OB at distance x and y from O then x + y = 12 and xy = 24.

x, y = 6 ± 2 (not possible)

So there is a unique line possible

Let Point P be ( α , β )

Using parametric equation AB

β = 6 – (6 + ) and α = (6 + )

⇒ Slope of PQ is = .