Prove that the points (a, 0), (0, b) and (1, 1) are collinear if `1/a+1/b=1`

#### Solution

The formula for the area ‘*A*’ encompassed by three points(x_{1},y_{1}),(x_{2},y_{2}) and (x_{3},y_{3}) is given by the formula,

We know area of triangle formed by three points (x_{1}y_{1}),(x_{2},y_{2}) and (x_{3},y_{3})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