Question

If the points (x1,y1) ,(x2,y2) and (x3,y3) be collinear show that y2-y3/x2x3 + y3-y1/x3x1 +y1-y2/x1x2 =0

Answer 1

We need to be familiar with the concepts of co-ordinate geometry to solve this question easily.
The steps cab be seen in the image.

Answer 2

(y₂ - y₃)/x₂x₃  + (y₃ - y₁)/x₁x₃  + (y₁ - y₂)/x₁x₂ = 0 if

(x₁ , y₁) , ( x₂ , y₂)  & (x₃ , y₃) are colinear Points

Step-by-step explanation:

(x₁ , y₁) , ( x₂ , y₂)  & (x₃ , y₃) are colinear Points

Hence area between them = 0

=> (1/2) |  x₁(y₂ - y₃)  + x₂(y₃ - y₁)  + x₃(y₁ - y₂) | = 0

Multiplying by 2 both sides

=>  |  x₁(y₂ - y₃)  + x₂(y₃ - y₁)  + x₃(y₁ - y₂) | = 0

Dividing both sides by x₁x₂x₃

=>  |  x₁(y₂ - y₃)/x₁x₂x₃  + x₂(y₃ - y₁)/x₁x₂x₃  + x₃(y₁ - y₂)/x₁x₂x₃ | = 0

=>  |  (y₂ - y₃)/x₂x₃  + (y₃ - y₁)/x₁x₃  + (y₁ - y₂)/x₁x₂ | = 0

=>  (y₂ - y₃)/x₂x₃  + (y₃ - y₁)/x₁x₃  + (y₁ - y₂)/x₁x₂ = 0

QED

Hence proved

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