Question

If ab+bc+ca=0,show that the lines x/a+y/b=1/c,x/b+y/c=1/a and x/c+y/a=1/b are concurrent?

Answer 1

See below.

Explanation:

If `ab+bc+ca=0`, show that the lines `x/a+y/b=1/c,x/b+y/c=1/a` and `x/c+y/a=1/b` are concurrent?

Given the Matrix

`M = ((1/a, 1/b, -1/c),(1/b, 1/c, -1/a),(1/c, 1/a, -1/b))`

if the three lines represented have a common point then their coefficients are linearly dependent and then

`det(M) = 1/a^3 + 1/b^3 + 1/c^3 - 3/(a b c)=0`

but

# 1/a^3 + 1/b^3 + 1/c^3 - 3/(a b c)=((a b + a c + b c) (a^2 b^2 - a^2 b c - a b^2 c + a^2 c^2 - a b c^2 + b^2 c^2))/(a^3b^3c^3)#

and as we can observe this relationship is true if

`a b + a c + b c=0`

Answer 2

Let the point of intersection of two straight lines `x/a+y/b=1/c .......[1]and x/b+y/c=1/a......[2]` be `(h,k)`.

So `(h,k)` will satisfy both the equations and we have the following two relations

`h/a+k/b=1/c .......[3]`

and

`h/b+k/c=1/a........[4]`

Adding [3] and [4] we get

`h(1/a+1/b)+k(1/b+1/c)=1/c+1/a.............[5]`

But the given condition is

`ab+bc+ca=0`

Dividing both sides by `abc` we get

`1/a+1/b+1/c=0...........[6]`

Combining {5}and [6] we get

`h(-1/c)+k(-1/a)=-1/b`

`=>h/c+k/a=1/b.....[7]`

The relation [7] is only possible iff `(h.k)`, the point of intersection of [1] and [2] also satisfies given 3rd equation `x/c+y/a=1/b`.

Hence we can say that the given three straight lines are concurrent under given condition `ab+bc+ca=0`