If points (a, 0), (0, b) and (1, 1) are collinear, then $\frac{a+b}{ab}$ is equal to :

Question

Accepted Answer

To determine why Option 1 is the correct answer, we need to verify the condition for collinearity of the points $(a, 0)$, $(0, b)$, and $(1, 1)$. Collinearity means that the area of the triangle formed by these three points is zero.

The area of a triangle formed by points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

$$
\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|
$$

Substituting the given points $(a, 0)$, $(0, b)$, and $(1, 1)$ into the formula, we have:

$$
\text{Area} = \frac{1}{2} \left| a(b-1) + 0(1-0) + 1(0-b) \right|
$$

Simplifying, we get:

$$
\text{Area} = \frac{1}{2} \left| ab - a - b \right|
$$

For the points to be collinear, the area must be zero:

$$
\left| ab - a - b \right| = 0
$$

This implies:

$$
ab - a - b = 0
$$

Rearranging the equation, we get:

$$
ab = a + b
$$

Dividing both sides by $ab$, we obtain:

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

Thus, $\frac{a+b}{ab} = 1$, which corresponds to Option 1.

Now, let's explain why the other options are incorrect:

- **Option 2 (-1):** If $\frac{a+b}{ab} = -1$, it would imply that $a + b = -ab$, which contradicts the derived equation $ab = a + b$. Therefore, Option 2 is incorrect.

- **Option 3 (2):** If $\frac{a+b}{ab} = 2$, it would imply that $a + b = 2ab$, which again contradicts the equation $ab = a + b$. Hence, Option 3 is incorrect.

- **Option 4 ($\sqrt{2}$):** If $\frac{a+b}{ab} = \sqrt{2}$, it would imply that $a + b = \sqrt{2}ab$, which does not satisfy the condition $ab = a + b$. Therefore, Option 4 is incorrect.

In conclusion, the only value that satisfies the condition for collinearity is $\frac{a+b}{ab} = 1$, making Option 1 the correct answer.