Question

Let O(0, 0), A(2, 0), B(1, 1√3) be the vertices of a triangle. Let R be the region consisting of all those points P inside triangle OAB, which satisfy d (P, OA) ≥ min{(P, OB), d(P, AB)} where d denotes the distance from the point to the corresponding line. Sketech the region R and find its area.

Answer 1

Let the co-ordinates of P be (x, y) 

Equation of the line OA: y = 0 

Equation of the line OB: √3y = x  

Equation of the line AB: √3y = 2 – x 

d(P, OA) = Distance of P from the line OA = y 

d(P, OB) = Distance of P from the line OB

= |R3y – x|/2

d (P, AB) = Distance of P from the line AB

= |R3y – x – 2|/2

It is given that

d (P, OA) ≤ min{d(P, OB), d (P,OA)}

Case I: When y ≤ |R3y – x|/2, then

Case II: When y ≤ |√3y – x + 2|/2, then

Hence, the area of the shaded region

= (1/2) × base × height

= (1/2) × 2 × (1.tan15°)

= tan 15° 

= (2 – √3) sq.u.