The points A ($x$, 0), B (−1, −1) and C (0, $y$) are collinear. The value of $\frac{1}{x} + \frac{1}{y}$ is:

Given,

The points A ($x$, 0), B (−1, −1) and C (0, $y$) are collinear.

We know that,

Collinear points lie on the same straight line, so they cannot enclose a triangle.

∴ Area of the triangle formed by them = 0.

By formula,

Area of triangle = $\frac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)]$

Here,

$$\begin{aligned}\Rightarrow x_1 = x, y_1 = 0 \\[1em] \Rightarrow x_2 = −1, y_2 = −1 \\[1em] \Rightarrow x_3 = 0, y_3 = y\end{aligned}$$

Substituting values we get :

$\Rightarrow \frac{1}{2}[x(-1 - y) + (-1)(y - 0) + 0(0 - (-1))] = 0$

Multiplying both sides by 2,

$$\begin{aligned}\Rightarrow x(-1 - y) + (-1)(y - 0) + 0(0 - (-1)) = 0 \\[1em] \Rightarrow -x - xy - y + 0 = 0 \\[1em] \Rightarrow -(x + y + xy) = 0 \\[1em] \Rightarrow x + y + xy = 0 \\[1em] \Rightarrow x + y = -xy.\end{aligned}$$

Since $\frac{1}{x} + \frac{1}{y}$ is required, we divide both sides by $xy$ (where $x ≠ 0$ and $y ≠ 0$),

$$\begin{aligned}\Rightarrow \frac{x + y}{xy} = \frac{-xy}{xy} \\[1em] \Rightarrow \frac{x}{xy} + \frac{y}{xy} = -1 \\[1em] \Rightarrow \frac{1}{y} + \frac{1}{x} = -1 \\[1em] \Rightarrow \frac{1}{x} + \frac{1}{y} = -1.\end{aligned}$$

Hence, option 1 is the correct option.