Question

A rectangle is formed by the lines \( x = 0,\ y = 0,\ x = 3,\ y = 4 \). A line perpendicular to \( 3x + 4y + 6 = 0 \) divides the rectangle into two equal parts. Then the distance of the line from the point \( \left(-1,\frac{3}{2}\right) \) is:

• A. \(2\)
• B. \( \frac{17}{10} \)
• C. \( \frac{6}{5} \)
• D. \( \frac{8}{5} \)

Hint:

Any line that divides a rectangle into two equal areas must pass through its centroid.

Answer 1

The Correct Option is B

Concept:
A line dividing a plane figure into two equal areas must pass through the centroid of the figure.
Centroid of a rectangle \( = \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \)
Distance of a point from a line \( ax+by+c=0 \) is \[ d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}} \]
Step 1: Find the centroid of the rectangle. The rectangle is bounded by: \[ 0 \le x \le 3,\quad 0 \le y \le 4 \] Hence, centroid: \[ \left(\frac{0+3}{2},\frac{0+4}{2}\right) = \left(\frac{3}{2},2\right) \]
Step 2: Find the slope of the required line. Given line: \[ 3x+4y+6=0 \Rightarrow y=-\frac{3}{4}x-\frac{3}{2} \] Slope \( m_1 = -\frac{3}{4} \) Slope of a perpendicular line: \[ m_2 = \frac{4}{3} \]
Step 3: Equation of the line passing through the centroid. Using point–slope form: \[ y-2=\frac{4}{3}\left(x-\frac{3}{2}\right) \] Simplifying: \[ y-2=\frac{4}{3}x-2 \Rightarrow y=\frac{4}{3}x \] Thus, the line is: \[ 4x-3y=0 \]
Step 4: Find the distance of the point \( \left(-1,\frac{3}{2}\right) \) from the line. \[ d=\frac{|4(-1)-3(\frac{3}{2})|}{\sqrt{4^2+(-3)^2}} \] \[ =\frac{| -4-\frac{9}{2}|}{\sqrt{25}} =\frac{\frac{17}{2}}{5} =\frac{17}{10} \]