Prove that the points $(a, b), (a_1, b_1)$ and $(a – a_1, b – b_1)$ are collinear if $ab_1 = a_1b$.

Given:

Given points are $(a, b), (a_1, b_1)$ and $(a – a_1, b – b_1)$.

To do:

We have to prove that the points $(a, b), (a_1, b_1)$ and $(a – a_1, b – b_1)$ are collinear if $ab_1 = a_1b$.

Solution:

Let $A(a, b), B(a_1, b_1)$ and $C(a-a_1, b-b_1)$ be the vertices of $\triangle ABC$.

We know that,

Area of a triangle with vertices $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ is given by, 

Area of $\Delta=\frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]$

Therefore,

Area of triangle \( ABC=\frac{1}{2}[a(b_1-b+b_1)+a_1(b-b_1-b)+(a-a_1)(b-b_1)] \)

\( =\frac{1}{2}[a(2b_1-b)+a(-b_1)+(a-a_1)(b-b_1)] \)

\( =\frac{1}{2}[2ab_1-ab-ab_1+ab-ab_1-a_1b+a_1b_1] \)

\( =\frac{1}{2}[ab_1-a_1b] \)

\( =\frac{1}{2}[ab_1-ab_1 \)     ($a_1b=ab_1$)

\( =0 \)

Therefore, points $A, B$ and $C$ are collinear.

Hence proved. 

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Updated on: 10-Oct-2022

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