Question

If \( \beta \) is the angle made by the perpendicular drawn from origin to the line \( L = x + y - 2 = 0 \) with the positive X-axis in the anticlockwise direction. If \( a \) is the X-intercept of the line \( L = 0 \) and \( p \) is the perpendicular distance from the origin to the line \( L = 0 \), then \( \tan \beta + p^2 = \):

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When dealing with problems that require finding intercepts, perpendicular distances, and slopes, remember to apply the standard formulas for each and verify the arithmetic at each step.

Answer 1

The Correct Option is C

Consider the line given by: \[ L: x + y - 2 = 0. \] This line is in the standard form \(Ax + By + C = 0\) with \(A = 1\), \(B = 1\), and \(C = -2\). 

 Step 1: Determine the X-intercept 
Set \(y = 0\) in the equation: \[ x + 0 - 2 = 0 \quad \Rightarrow \quad x = 2. \] Thus, the X-intercept is \(a = 2\). 

Step 2: Find the Perpendicular Distance from the Origin 
The perpendicular distance \(p\) from a point \((x_1, y_1)\) (here, the origin \((0,0)\)) to the line \(Ax + By + C = 0\) is given by: \[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}. \] Substituting \(A = 1\), \(B = 1\), \(C = -2\) and \((x_1, y_1) = (0,0)\): \[ p = \frac{|1(0) + 1(0) - 2|}{\sqrt{1^2 + 1^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}. \] 

Step 3: Determine \(\tan \beta\) 
Here, \(\beta\) is the angle that the line makes with the positive X-axis. Since the line \(x+y-2=0\) can be rewritten as \(y = 2 - x\), its slope is: \[ m = -1. \] Thus, the absolute value of the slope gives: \[ \tan \beta = |m| = 1. \] 

Step 4: Compute \(\tan \beta + p^2\) 
We have: \[ \tan \beta = 1 \quad \text{and} \quad p^2 = (\sqrt{2})^2 = 2. \] Therefore: \[ \tan \beta + p^2 = 1 + 2 = 3. \] \bigskip However, as per the final statement, the answer provided is: \[ \boxed{3}. \]