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Let O(0,0),A(2,0),a n dB(1, 1/(sqrt(3)))...

Let `O(0,0),A(2,0),a n dB(1, 1/(sqrt(3)))` be the vertices of a triangle. Let `R` be the region consisting of all those points `P` inside ` O A B` which satisfy `d(P , O A)lt=min[d(P, O B),d(P,A B)]` , where `d` denotes the distance from the point to the corresponding line. Sketch the region `R` and find its area.

A

`sqrt(3)`sq,units

B

`(2+sqrt3)` sq.units

C

`sqrt(3)//2` sq.units

D

`1//sqrt(3)` sq.units

Text Solution

Verified by Experts

The correct Answer is:
D

`d(P,OA)le min[d(P,OB),d(P,AB)]`
or `d(P,OA)le(P,OB)`
and `d(P,OA0)le d(P,AB)`
When `d(P,OA)P` is equidistant from OA and OB, or P lies on the bisector of lines OA and OB.
Hence, when `d(P,OA)le d(P,OB)`, point P is nearer to OA than to OB or lies below the angle bisector of `angleAOB`. Similarly, when `d(P,OA)led(P,AB)P` is nearer to OA than to AB, or P lies below the bisector of `angle OAB and AB`. Therefore, the required area is equal to the area of `DeltaOIA`. Now,
`tanangleBOA=(sqrt3)/(1)=sqrt3`
or `angle BOA=60^@`
Hence, the triangle is equilateral. then I coincides. with the centroid which is `(1,1//sqrt3)`.
Therefore,the area of `Delta OIA` is
`(1)/(2) OAxxIM=(1)/(2)xx2xx(1)/(sqrt3)=(1)/(sqrt3)`sq.units
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