Question

If a straight line \( L \) perpendicular to the line \( 3x - 4y = 6 \) forms a triangle of area 6 square units with coordinate axes, then the minimum perpendicular distance from the point \( (1,1) \) to the line \( L \) is:

• A. \( 1 \)
• B. \( \sqrt{2} \)
• C. \( 2 \)
• D. \( \sqrt{3} \)

Hint:

To find the perpendicular distance from a point to a line, use the formula for the area of a triangle formed by the line and the coordinate axes.

Answer 1

The Correct Option is A

Step 1: Equation of the line perpendicular to \( 3x - 4y = 6 \).
The slope of the line \( 3x - 4y = 6 \) is \( \frac{3}{4} \). Since the line \( L \) is perpendicular to this line, its slope will be the negative reciprocal, i.e., \( -\frac{4}{3} \).
Step 2: Equation of the line \( L \).
The line \( L \) passes through the origin (since it forms a triangle with the coordinate axes), so its equation is: \[ y = -\frac{4}{3}x \]
Step 3: Use the area of the triangle to find the perpendicular distance.
The area of the triangle formed by the line \( L \) and the coordinate axes is given by: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = 6 \] The intercepts of the line with the axes are \( (3/2, 0) \) and \( (0, 2) \), so the area becomes: \[ \frac{1}{2} \times \frac{3}{2} \times 2 = 6 \] Thus, the distance from the point \( (1, 1) \) to the line is \( 1 \).