The foot of the perpendicular on the line `3x+y=lambda` drawn from the origin is `Cdot` If the line cuts the `x` and the y-axis at `Aa n dB` , respectively, then `B C: C A` is

Options

A. `1:3`
B. `3:1`
C. `1:9`
D. `9:1`

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the Problem
We need to find the ratio \( BC : CA \) where \( C \) is the foot of the perpendicular from the origin \( O(0, 0) \) to the line given by the equation \( 3x + y = \lambda \). The line intersects the x-axis at point \( A \) and the y-axis at point \( B \).

### Step 2: Find the Points A and B
To find the points where the line intersects the axes:
- **Point A (x-intercept)**: Set \( y = 0 \) in the equation \( 3x + y = \lambda \):
\[ 3x + 0 = \lambda \implies x = \frac{\lambda}{3} \]
So, \( A\left(\frac{\lambda}{3}, 0\right) \).

- **Point B (y-intercept)**: Set \( x = 0 \) in the equation \( 3x + y = \lambda \):
\[ 3(0) + y = \lambda \implies y = \lambda \]
So, \( B(0, \lambda) \).

### Step 3: Find the Slope of Line AB
The slope \( m \) of line \( AB \) can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\lambda - 0}{0 - \frac{\lambda}{3}} = \frac{\lambda}{-\frac{\lambda}{3}} = -3 \]

### Step 4: Determine the Angle \( \theta \)
From the slope, we can determine the angle \( \theta \) that the line makes with the positive x-axis:
\[ \tan \theta = -3 \implies \tan(180^\circ - \theta) = 3 \implies \tan \theta = 3 \]

### Step 5: Set Up the Right Triangles
In triangle \( OCA \):
- The angle at \( O \) is \( \theta \).
- The angle at \( C \) is \( 90^\circ \).
- We can express \( \tan \theta \) as:
\[ \tan \theta = \frac{OC}{AC} \quad \text{(Equation 1)} \]

In triangle \( OCB \):
- The angle at \( O \) is \( 180^\circ - \theta \).
- The angle at \( C \) is \( 90^\circ \).
- We can express \( \cot \theta \) as:
\[ \cot \theta = \frac{OC}{BC} \quad \text{(Equation 2)} \]

### Step 6: Relate the Two Equations
Dividing Equation 1 by Equation 2:
\[ \frac{OC}{AC} \div \frac{OC}{BC} = \frac{\tan \theta}{\cot \theta} \]
This simplifies to:
\[ \frac{BC}{AC} = \tan \theta \cdot \cot \theta = \tan^2 \theta \]
Since \( \tan \theta = 3 \):
\[ \frac{BC}{AC} = 3 \cdot \frac{1}{3} = 9 \]

### Step 7: Conclusion
Thus, the ratio \( BC : CA \) is:
\[ BC : CA = 9 : 1 \]