The line `4x+y=lamda` cuts the axes of co ordinates at A and B. If C is the foot of perendicular drawn from origin O, then AC:CB=

Options

A. `1:16`
B. `16:1`
C. `2:1`
D. `1:2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the points A and B where the line intersects the axes.
The line given is \(4x + y = \lambda\).

- **To find point A (x-intercept)**, set \(y = 0\):
\[ 4x + 0 = \lambda \implies x = \frac{\lambda}{4} \]
Thus, point A is \(\left(\frac{\lambda}{4}, 0\right)\).

- **To find point B (y-intercept)**, set \(x = 0\):
\[ 4(0) + y = \lambda \implies y = \lambda \]
Thus, point B is \((0, \lambda)\).

### Step 2: Find the coordinates of point C, the foot of the perpendicular from the origin O to the line.
The slope of the line \(4x + y = \lambda\) can be rearranged to the slope-intercept form:
\[ y = -4x + \lambda \]
The slope (m) of this line is \(-4\).

The slope of the perpendicular line from the origin O (0, 0) to this line will be the negative reciprocal:
\[ m_{OC} = \frac{1}{4} \]

The equation of line OC (from origin) is:
\[ y = \frac{1}{4}x \]

### Step 3: Find the intersection point C of lines OC and AB.
Substituting \(y = \frac{1}{4}x\) into the line equation \(4x + y = \lambda\):
\[ 4x + \frac{1}{4}x = \lambda \]
Multiplying through by 4 to eliminate the fraction:
\[ 16x + x = 4\lambda \implies 17x = 4\lambda \implies x = \frac{4\lambda}{17} \]
Now substituting \(x\) back to find \(y\):
\[ y = \frac{1}{4} \cdot \frac{4\lambda}{17} = \frac{\lambda}{17} \]
Thus, point C is \(\left(\frac{4\lambda}{17}, \frac{\lambda}{17}\right)\).

### Step 4: Calculate lengths AC and CB.
- **Length AC**:
\[ AC = \sqrt{\left(\frac{4\lambda}{17} - \frac{\lambda}{4}\right)^2 + \left(\frac{\lambda}{17} - 0\right)^2} \]
First, simplify \(x\) coordinate:
\[ \frac{4\lambda}{17} - \frac{\lambda}{4} = \frac{16\lambda - 17\lambda}{68} = \frac{-\lambda}{68} \]
Thus,
\[ AC = \sqrt{\left(\frac{-\lambda}{68}\right)^2 + \left(\frac{\lambda}{17}\right)^2} = \sqrt{\frac{\lambda^2}{4624} + \frac{4\lambda^2}{4624}} = \sqrt{\frac{5\lambda^2}{4624}} = \frac{\lambda\sqrt{5}}{68} \]

- **Length CB**:
\[ CB = \sqrt{\left(0 - \frac{4\lambda}{17}\right)^2 + \left(\lambda - \frac{\lambda}{17}\right)^2} \]
Simplifying \(y\) coordinate:
\[ \lambda - \frac{\lambda}{17} = \frac{17\lambda - \lambda}{17} = \frac{16\lambda}{17} \]
Thus,
\[ CB = \sqrt{\left(-\frac{4\lambda}{17}\right)^2 + \left(\frac{16\lambda}{17}\right)^2} = \sqrt{\frac{16\lambda^2}{289} + \frac{256\lambda^2}{289}} = \sqrt{\frac{272\lambda^2}{289}} = \frac{\lambda\sqrt{272}}{17} \]

### Step 5: Find the ratio \(AC:CB\).
Now we can find the ratio:
\[ \frac{AC}{CB} = \frac{\frac{\lambda\sqrt{5}}{68}}{\frac{\lambda\sqrt{272}}{17}} = \frac{\sqrt{5}}{68} \cdot \frac{17}{\sqrt{272}} = \frac{17\sqrt{5}}{68\sqrt{272}} = \frac{17\sqrt{5}}{68 \cdot 16.49} \approx \frac{1}{16} \]
Thus, the ratio \(AC:CB\) simplifies to:
\[ 1:16 \]

### Final Answer:
The ratio \(AC:CB\) is \(1:16\).