ABC is an equilateral triangle where AB = a and P is any point in its plane such that PA = PB + PC. Then `(PA^(2)+PB^(2)+PC^(2))/(a^(2))` is

Options

A. 3
B. `(sqrt(3))/(4)`
C. `(3)/(4)`
D. 2

Answer

D

Explanation

PA = PB + PC and ABC is equiliateral
`therefore` P lies on arc BC of circumcircle of `Delta ABC`

`therefore angle BPC = 120^(@)`
Using cosine rule in `Delta BPC`
`cos 120^(@)=(PB^(2)+PC^(2)-BC^(2))/(2PB.PC)`
`rArr -PB.PC=PB^(2)+PC^(2)-a^(2)`
`rArr 2PB.PC=2a^(2)-2(PB^(2)+PC^(2))`
Again, PA = PB + PC
`rArr PA^(2)=PB^(2)+PC^(2)+2PB.PC`
`rArr PA^(2)=PB^(2)+PC^(2)+2a^(2)-2(PB^(2)+PC^(2))`
`(PA^(2)+PB^(2)+PC^(2))/(a^(2))=2`