Question

If the points (a, 0), (0, b) and (1, 1) are collinear, then $\frac{1}{a}+\frac{1}{b}$ = ___________.

Answer 1

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}$ $\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$.

Now, the given points (a, 0), (0, b) and (1, 1) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=0 \\ \Rightarrow \frac{1}{2}\left|a\left(b-1\right)+\left(0\right)\left(1-0\right)+1\left(0-b\right)\right|=0 \\ \Rightarrow \frac{1}{2}\left|ab-a+0-b\right|=0 \\ \Rightarrow \left|ab-a-b\right|=0 \\ \Rightarrow ab-a-b=0 \\ \Rightarrow ab=a+b \\ \Rightarrow a+b=ab \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}ab,\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{a}{ab}+\frac{b}{ab}=\frac{ab}{ab} \\ \Rightarrow \frac{1}{b}+\frac{1}{a}=1 \end{gathered}$$

Hence, $\frac{1}{a}+\frac{1}{b}$ = 1.