Question

(a,b),(b,a) and (1/a,1/b) these point are in one line. now proved that a+b=0​

Answer 1

solution

rule:- if three points are in one line then area of

triangle formed by them = 0

so

Delta= 1/2{ (a×a+ b×1/b+ 1/a×b ) - ( b×b+a×1/a+1/b×a)}

= 1/2{( a²+1+b/a)- (b²+1+a/b)}= 0

= 1/2( a²-b²+b/a-a/b)= 0

-( a²-b²) ab+b²-a²==0

- a³b+ab³+b²-a²==0

ab³+b²- a³b-a²=0

b²( ab+1) - a²( ab+1)=0

(ab+1) ( b2-a²)= 0

b²-a²=0

( b-a) (b+a)= 0

a+b=0

proved