In a triangle ABC, AB = 5 , BC = 7, AC = 6. A point P is in the plane such that it is at distance '2' units from AB and 3 units form AC then prove its distance from BC is `(12 sqrt6-28)/(7)` when P is inside the trinagle

Answer

A, B, C

Explanation

`a = 7, b = 6, c = 5`
`:. s = 9`
`:. "Area " Delta = sqrt(s(s-a) (s-b) (s-c)) = 6sqrt6`
Now, PF = 2, PE = x
Let PD = x
Then
`Delta = (1)/(2) xx 2 xx 5 + (1)/(2) xx x xx 7 + (1)/(2) xx 3 xx 6`
`:. 6sqrt6 = 5 +(7x)/(2) + 9`
`:. x = (12 sqrt6 -28)/(7)`
When P lies outside the triangle

Area of `DeltaABC = (1)/(2) x xx 7 + (1)/(2) xx 3 xx 6 - (1)/(2) xx 2 xx 5`
`:. 6sqrt6 = (7)/(2) x + 9 -5`
`:. x = (12 sqrt6 -8)/(7)`
Similarly, in one more case, we get `(12 sqrt6 +8)/(7)`