Question

A light ray emits from the origin making an angle \(30\degree\) 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

• A. 2/3-√3
• B. 2/3+√3
• C. √3/2(√3+1)
• D. 2/(√3-1)

Answer 1

The Correct Option is B

Step 1: Finding the slope of the reflected ray. 
The light ray emits from the origin and makes an angle of \(30^\circ\) with the positive x-axis. The slope of the reflected ray is: \[ \text{slope of reflected ray} = \tan 60^\circ = \sqrt{3}. \] Step 2: Finding the equation of the reflected ray. The equation of the reflected ray will be in the form: \[ y = m x + c, \] where \(m = \sqrt{3}\) is the slope and \(c\) is the y-intercept. Since the ray passes through the origin, \(c = 0\). Therefore, the equation of the reflected ray is: \[ y = \sqrt{3}x. \] Step 3: Finding the intersection point with the line \(x + y = 1\). 
The line \(x + y = 1\) intersects the reflected ray at the point where the equation \(y = \sqrt{3}x\) satisfies the equation \(x + y = 1\). Substituting \(y = \sqrt{3}x\) into the line equation: \[ x + \sqrt{3}x = 1 \quad \Rightarrow \quad x(1 + \sqrt{3}) = 1 \quad \Rightarrow \quad x = \frac{1}{1 + \sqrt{3}}. \] Step 4: Finding the abscissa of \(Q\). 
The abscissa of \(Q\) is found by substituting \(x = \frac{1}{1 + \sqrt{3}}\) into the equation of the reflected ray \(y = \sqrt{3}x\). The value of \(x\) gives the abscissa of the point where the ray intersects the x-axis. Rationalizing the denominator, we get: \[ x = \frac{1}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{1 - \sqrt{3}}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{1 - \sqrt{3}}{1 - 3} = \frac{1 - \sqrt{3}}{-2} = \frac{2}{3 + \sqrt{3}}. \] Thus, the abscissa of \(Q\) is \( \frac{2}{3 + \sqrt{3}} \), and the correct answer is option (2).