Answer 1

The formula for the area ‘A’ encompassed by three points(x₁,y₁),(x₂,y₂) and (x₃,y₃) is given by the formula,

We know area of triangle formed by three points (x₁y₁),(x₂,y₂) and (x₃,y₃)is given by 

Δ`=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

If three points are collinear the area encompassed by them is equal to 0.

The three given points are A(a,0), B(0,b) and C(1,1). 

A`=1/2[a(b-1)+1(0-b)]`

`=1/2[ab-a-b] ` 

It is given that `1/a+1/b=1`  

So we have, 

`1/a+1/b=1` 

`(a+b)/(ab)=1` 

a+b=ab 

Using this in the previously arrived equation for area we have, 

A`=ab-(a+b)`

A `ab-ab`

A=0 

Since the area enclosed by the three points is equal to 0, the three points need to be collinear