Question

A rectangle is formed by lines \( x = 0, y = 0, x = 3 \) and \( y = 4 \). 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. \( \frac{17}{10} \)
• B. \( \frac{10}{17} \)
• C. \( \frac{15}{17} \)
• D. \( \frac{18}{17} \)

Hint:

To find the distance from a point to a line, use the distance formula and substitute the coordinates of the point and the coefficients from the equation of the line.

Answer 1

The Correct Option is A

To solve this problem, we need to do the following steps:

1. Identify the equation of the line perpendicular to the given line \(3x + 4y + 6 = 0\).
2. Find the line that divides the rectangle into two equal areas.
3. Calculate the distance from the given point \( \left( -1, \frac{3}{2} \right) \) to that line.

Step 1: Find the Slope of the Given Line

The line is given by \(3x + 4y + 6 = 0\). Rearranging the terms yields:

\(4y = -3x - 6 \Rightarrow y = -\frac{3}{4}x - \frac{3}{2}\)

Thus, the slope of this line is \(-\frac{3}{4}\). A line perpendicular to this should have a slope that is the negative reciprocal: \(\frac{4}{3}\).

Step 2: Determine the Equation of the Perpendicular Line

The equation of a line with slope \(\frac{4}{3}\) is:

\(y = \frac{4}{3}x + c\)

Step 3: Divide the Rectangle Into Two Equal Areas

The rectangle is bounded by: \(x = 0\), \(y = 0\), \(x = 3\), \(y = 4\). Its area is \(3 \times 4 = 12\).

We need the line to divide this into two regions of area \(6\) each. The line must pass through the center of the rectangle \(\left(\frac{3}{2}, 2\right)\). Substituting this point into the line equation:

\(2 = \frac{4}{3} \cdot \frac{3}{2} + c \Rightarrow 2 = 2 + c \Rightarrow c = 0\)

Thus, the equation of the line is:

\(y = \frac{4}{3}x\). This line, through the center, divides the rectangle into two equal areas.

Step 4: Calculate the Distance From the Point to the Line

The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:

\(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

For our line \(y - \frac{4}{3}x = 0\), this can be rewritten as:

\(- \frac{4}{3}x + y = 0 \Rightarrow A = -\frac{4}{3}, B = 1, C = 0\)

Plugging in the point \(\left(-1, \frac{3}{2}\right)\):

\(d = \frac{\left| -\frac{4}{3} \cdot (-1) + 1 \cdot \frac{3}{2} + 0 \right|}{\sqrt{\left(-\frac{4}{3}\right)^2 + 1^2}}\)

\(d = \frac{\left| \frac{4}{3} + \frac{3}{2} \right|}{\sqrt{\frac{16}{9} + 1}}\)

Calculate the inside terms:

\(\frac{4}{3} + \frac{3}{2} = \frac{8}{6} + \frac{9}{6} = \frac{17}{6}\)

\(\sqrt{\frac{25}{9}} = \frac{5}{3}\)

Thus the distance is:

\(d = \frac{\frac{17}{6}}{\frac{5}{3}} = \frac{17}{6} \cdot \frac{3}{5} = \frac{17}{10}\)

Therefore, the distance of the line from the point \(\left(-1, \frac{3}{2} \right)\) is \(\frac{17}{10}\).