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

Question

Prove that the points (a, 0), (0, b) and (1, 1) are collinear if,\(\frac{1}a+\frac{1}b\) = 1

Answer 1

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) 

= \(\frac{1}2\) |x₁(y₂ - y₃)+x₂(y₃ - y₁)+x₃(y₁ - y₂)| 

Area of ∆ABC 

= \(\frac{1}2\) |a(b – 1) + 1(0 -b)| 

= \(\frac{1}2\) | ab – a –b| 

Here given that \(\frac{1}a+\frac{1}b\) = 1

∴ \(\frac{a+b}{ab}\) = 1

∴ a + b = ab 

Now, 

Area of ∆ABC 

= \(\frac{1}2\) | ab - (a + b)| 

= \(\frac{1}2\) | ab – ab| 

= \(\frac{1}2\) | 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.