The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle $${\pi \over 4}$$ at the origin, is equal to :

Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : $$-$$4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :

In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($$\alpha$$, $$\beta$$) is the centroid of $$\Delta$$ABC, then 15($$\alpha$$ + $$\beta$$) is equal to :

Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of $$\Delta$$PQR is :