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?

2 Answers
Aug 10, 2017

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#

Aug 11, 2017

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#