Question

If the points $(a,0)$, $(0,b)$ and $(1,1)$ are collinear then which of the following is true?

• A. $\frac{1}{a}+\frac{1}{b}=1$
• B. $\frac{1}{a}-\frac{1}{b}=2$
• C. $\frac{1}{a}-\frac{1}{b}=-1$
• D. $\frac{1}{a}+\frac{1}{b}=2$

Answer 1

The correct option is A

$\frac{1}{a}+\frac{1}{b}=1$

Explanation of correct option:

Step 1: Calculate the area of triangle.

The given points are $(a,0)$, $(0,b)$ and $(1,1)$ are collinear.

Now, we know that the area $\left(A\right)$ of a triangle with vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ can be calculated by the following formula,

$A=\frac{1}{2}\left|\begin{array}{ccc}1& x_1& y_1\\ 1& x_2& y_2\\ 1& x_3& y_3\end{array}\right|$

Here $\left(x_1,y_1\right),\left(x_2,y_2\right)$and $\left(x_3,y_3\right)$are$\left(a,0\right),\left(0,b\right)$ and $(1,1)$ respectively.

So, $A=\frac{1}{2}\left|\begin{array}{ccc}1& a& 0\\ 1& 0& b\\ 1& 1& 1\end{array}\right|$

$\Rightarrow A=\frac{1}{2}\left[a(b-1)+0(1-0)+1(0-b)\right]$ [simplifying the matrix]

$\Rightarrow A=\frac{1}{2}\left[ab-a-b\right]$

Step 2: Find the relation between $a$and $b$

Now, the area of the triangle with vertices $\left(a,0\right),\left(0,b\right)$ and $(1,1)$ is zero as the given points $(a,0)$, $(0,b)$ and $(1,1)$ are collinear.

i.e., $A=0$

$\Rightarrow$ $\frac{1}{2}\left[ab-a-b\right]=0$

$\Rightarrow$ $ab-a-b=0$

$\Rightarrow$ $a+b=ab$

$\Rightarrow$ $\frac{a+b}{ab}=\frac{ab}{ab}$

$\Rightarrow$ $\frac{1}{a}+\frac{1}{b}=1$

Hence, option A is the correct option.