Question

The number of straight lines that can be drawn through the point $ (-3, 4) $, which are at a distance of 5 units from the point $ (2, -8) $, is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( \text{infinite} \)

Hint:

The distance from a point \( (x_0, y_0) \) to a line \( ax + by + c = 0 \) is \( \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \). Check vertical lines separately.

Answer 1

The Correct Option is C

Step 1: Set up the line equation. 
Line through \( (-3, 4) \): \( y - 4 = m(x + 3) \), or \( mx - y + 3m + 4 = 0 \). 
Step 2: Apply the distance condition. 
Distance from \( (2, -8) \): \( \frac{|m \cdot 2 - (-8) + 3m + 4|}{\sqrt{m^2 + 1}} = \frac{|5m + 12|}{\sqrt{m^2 + 1}} = 5 \). 
\[ (5m + 12)^2 = 25 (m^2 + 1) \quad \Rightarrow \quad 120m + 119 = 0 \quad \Rightarrow \quad m = -\frac{119}{120}. \] One line: \( y - 4 = -\frac{119}{120}(x + 3) \). 
Step 3: Check the vertical line. 
Line \( x = -3 \). Distance from \( (2, -8) \): \( |2 - (-3)| = 5 \). Second line. 
Step 4: Conclusion. 
Total lines = 2.