Prove that the points (a, 0), (0, b) and (1, 1) are collinear if,

Let three given points be A(a,0), B(0,b) and C(1,1).

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆ABC

= |a(b – 1) + 1(0 -b)|

= | ab – a –b|

Here given that

∴ = 1

∴ a + b = ab

Now,

Area of ∆ABC

= | ab - (a + b)|

= | ab – ab|

= | 0 |

= 0 sq. units

We know that if area enclosed by three points is zero, then points are collinear.

Hence, given three points are collinear.