A light ray emits from the origin making angle $$30°$$ with the positive $$x$$-axis. After getting reflected by the line $$x + y = 1$$, if this ray intersects x-axis at Q, then the abscissa of Q is

Let the incident ray emit from the origin $$O(0, 0)$$ making an angle of $$30^\circ$$ with the positive $$x$$-axis. Its equation is given by:

$$y = \tan(30^\circ)x = \frac{1}{\sqrt{3}}x \implies x = \sqrt{3}y$$

The reflecting line acts as a plane mirror:

$$x + y = 1$$

According to the laws of reflection, the line containing the reflected ray will pass through the mirror image of the source point with respect to the line.

Step 1: Find the image of the origin $$O(0, 0)$$ in the line $$x + y - 1 = 0$$

Using the standard mirror image formula:

$$\frac{x_2 - 0}{1} = \frac{y_2 - 0}{1} = -2\frac{1(0) + 1(0) - 1}{1^2 + 1^2} = -2\left(\frac{-1}{2}\right) = 1$$

This gives $$x_2 = 1$$ and $$y_2 = 1$$. Thus, the image of the origin is the point $$M(1, 1)$$, and the line containing the reflected ray must pass through $$M$$.

Step 2: Find the point of incidence $$P$$

The incident ray intersects the line $$x + y = 1$$ at the point of incidence $$P$$. Substituting $$x = \sqrt{3}y$$ into the mirror line equation:

$$\sqrt{3}y + y = 1 \implies y(\sqrt{3} + 1) = 1 \implies y = \frac{1}{\sqrt{3} + 1}$$

Substituting $$y$$ back to find the $$x$$-coordinate of $$P$$:

$$x = \frac{\sqrt{3}}{\sqrt{3} + 1}$$

So, the point of incidence is $$P\left(\frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1}\right)$$.

Step 3: Find the slope of the reflected ray ($$PM$$)

The reflected ray passes through both the point of incidence $$P$$ and the image point $$M(1, 1)$$. The slope $$m$$ of the line $$PM$$ is:

$$m = \frac{1 - \frac{1}{\sqrt{3} + 1}}{1 - \frac{\sqrt{3}}{\sqrt{3} + 1}} = \frac{\frac{\sqrt{3} + 1 - 1}{\sqrt{3} + 1}}{\frac{\sqrt{3} + 1 - \sqrt{3}}{\sqrt{3} + 1}} = \frac{\sqrt{3}}{1} = \sqrt{3}$$

Step 4: Find the equation of the reflected ray

Using the point-slope form with $$M(1, 1)$$ and slope $$m = \sqrt{3}$$:

$$y - 1 = \sqrt{3}(x - 1)$$

Step 5: Find the intersection point $$Q$$ on the $$x$$-axis

The ray intersects the $$x$$-axis at $$Q$$, where $$y = 0$$. Substituting $$y = 0$$ into the reflected ray equation:

$$0 - 1 = \sqrt{3}(x - 1)$$

$$-1 = \sqrt{3}x - \sqrt{3}$$

$$\sqrt{3}x = \sqrt{3} - 1$$

$$x = \frac{\sqrt{3} - 1}{\sqrt{3}} = 1 - \frac{1}{\sqrt{3}}$$

To match standard rationalised formats, multiplying the numerator and denominator by $$(\sqrt{3} + 1)$$:

$$x = \frac{(\sqrt{3} - 1)(\sqrt{3} + 1)}{\sqrt{3}(\sqrt{3} + 1)} = \frac{3 - 1}{3 + \sqrt{3}} = \frac{2}{3 + \sqrt{3}}$$

Conclusion:

The abscissa of Q is equal to $$\frac{2}{3 + \sqrt{3}}$$ .