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Prove that the points $(a, 0), (0, b)$ and $(1, 1)$ are collinear if, $\frac{1}{a} + \frac{1}{b} = 1$.
Given:
Given points are $(a, 0), (0, b)$ and $(1, 1)$.
$\frac{1}{a} + \frac{1}{b} = 1$.
To do:
We have to prove that the given points are collinear.
Solution:
Let $A(a, b), B(a_1, b_1)$ and $C(a-a_1, b-b_1)$ be the vertices of $\triangle ABC$.
$\frac{1}{a} + \frac{1}{b} = 1$
$\Rightarrow \frac{b+a}{ab}=1$
$\Rightarrow a+b=1(ab)$
$\Rightarrow ab-a-b=0$.......(i)
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 \( ABC=\frac{1}{2}[a(b-1)+0(1-0)+1(0-b)] \)
\( =\frac{1}{2}[ab-a+0-b] \)
\( =\frac{1}{2}[ab-a-b] \)
\( =0\) (From (i))
Therefore, points $A, B$ and $C$ are collinear.
Hence proved.
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