Question

Consider a triangle OAB formed by the points O = (0, 0), A = (2, 0), B = (1, $\sqrt{3}$). P(x, y) is an arbitrary interior point of the triangle, moving in such a way that d(P, OA) + d(P, PB) + d(P, OB) = $\sqrt{3}$, where d(P, OA), d(P, AB), d(P, OB) represent the distance of 'P' from the sides OA, AB and OB respectively. Area of the region representing all possible positions of the point 'P' is equal to

• A. $2\sqrt{3}$sq. units
• B. $\sqrt{6}$sq. units
• C. $\sqrt{3}$sq. units
• D. None of these

Answer 1

Answer: (C)

$\sqrt{3}$sq. units

Answer 2

Triangle OAB is clearly equilateral.

∆_OAB ⁼ ∆_OPA + ∆_APB + ∆_OPB

$\Rightarrow \frac { \sqrt { 3 } } { 4 } \cdot 4 = \frac { 1 } { 2 }$ . 2. (d(P, OA) + d(P, AB) + d(P, OB))

⇒ d(P, OA) + d(P, AB) + d(P, OB) = $\sqrt { 3 }$

Thus, point 'P' can be any interior point of the triangle. Hence area of required region

= ∆_0AB = $\sqrt { 3 }$ sq. units.