If non-zero numbers a, b, c are in H.P., then the straight linex/a + y/b +1/c = 0 alwayspasses through a fixed point. That point is ?

Hello student,

recall that a , b ,c in harmonic progression then
b = a / ( 1 +d ) , c = a / (1 +2*d )
from b =a /(1+d ) solve for d = (a-b) /b
plug in then c = (ab) / (2a-b)
the equation of the line becomes
x/ a +y /b + (2a-b) / ab
or (x-1)/a + (y+2) /b =0 (1)
If each term is zero then equation (1) always holds.
So:
x- 1 =0
y +2 =0
Or x =1 , y= -2
Or the line always passing the fixed point ( x= 1 ,y =-2 ) .

Thanks and Regards

Shaik Aasif

askIITians faculty

Since abc are in HP so 

b = 2ac/ a+c

ab + bc = 2ac

Multiply by 1/abc both side

ab/abc + bc/abc - 2ac/abc =0

We get

1/a -2/b +1/c = o

So points are x= 1,y= -2

(1,-2)