Question

If \( M \) is the foot of the perpendicular drawn from the origin to the line \( x - 2y + 3 = 0 \), which meets the X and Y-axes at \( A \) and \( B \) respectively, then \( AM = \)

• A. \( \dfrac{6\sqrt{10}}{5} \)
• B. \( 6\sqrt{5} \)
• C. \( \dfrac{6\sqrt{5}}{5} \)
• D. \( \sqrt{10} \)

Hint:

Use the formula for the foot of the perpendicular from a point to a line to get the coordinates, then apply distance formula to find required lengths.

Answer 1

The Correct Option is C

Step 1: Find coordinates of A and B 
Given line: \( x - 2y + 3 = 0 \) To find X-intercept \( A \), set \( y = 0 \): \[ x + 3 = 0 \Rightarrow x = -3 \Rightarrow A = (-3, 0) \] To find Y-intercept \( B \), set \( x = 0 \): \[ -2y + 3 = 0 \Rightarrow y = \frac{3}{2} \Rightarrow B = (0, \tfrac{3}{2}) \] Step 2: Find coordinates of M 
Foot of perpendicular from origin \((0, 0)\) to line \( x - 2y + 3 = 0 \) is given by: 

Alternate method (shortcut): Foot of perpendicular from point \( (x_0, y_0) \) to line \( ax + by + c = 0 \) is: \[ M = \left( x_0 - a \cdot \frac{ax_0 + by_0 + c}{a^2 + b^2}, \ y_0 - b \cdot \frac{ax_0 + by_0 + c}{a^2 + b^2} \right) \] Here: \( a = 1, b = -2, c = 3 \) and \( (x_0, y_0) = (0, 0) \) \[ \Rightarrow M = \left( -1 \cdot \frac{0 + 0 + 3}{1^2 + (-2)^2},\ -(-2) \cdot \frac{3}{1^2 + 4} \right) = \left( -\frac{3}{5}, \frac{6}{5} \right) \] Step 3: Find distance \( AM \) 
A is \( (-3, 0) \), M is \( \left(-\frac{3}{5}, \frac{6}{5} \right) \) \[ AM = \sqrt{ \left( -\frac{3}{5} + 3 \right)^2 + \left( \frac{6}{5} - 0 \right)^2 } = \sqrt{ \left( \frac{12}{5} \right)^2 + \left( \frac{6}{5} \right)^2 } = \sqrt{ \frac{144 + 36}{25} } = \sqrt{ \frac{180}{25} } = \sqrt{7.2} = \frac{\sqrt{180}}{5} = \frac{6\sqrt{5}}{5} \] % Final Result \[ \boxed{AM = \dfrac{6\sqrt{5}}{5}} \]