A rectangle is formed by the lines x = 0, y = 0, x = 3 and y = 4. Let the line L be perpendicular to

Q: Why does the line have to pass through the center of the rectangle?

Think of the rectangle as a perfectly balanced object. Any line passing through the center of a rectangle creates two congruent trapezoids (or rectangles), which by symmetry must have equal areas. It is a fundamental property of central symmetry in geometry.

Q: Why is the slope of the perpendicular line the negative reciprocal?

If a line has slope $m$, a perpendicular line must have a slope $m'$ such that $m \cdot m' = -1$. This comes from the geometric relationship between the angles of inclination of perpendicular lines, which differ by $90^\circ$. Since $\tan(\theta + 90^\circ) = -\cot(\theta) = -\frac{1}{\tan(\theta)}$, the negative reciprocal is the only way to satisfy the condition.

Q: Why do we use the absolute value in the distance formula?

The distance between a point and a line must always be a non-negative scalar value. The expression $Ax_1 + By_1 + C$ can be positive or negative depending on which side of the line the point lies. The absolute value ensures we are measuring the magnitude of the separation, not the signed displacement.

Question

A rectangle is formed by the lines $x = 0, y = 0, x = 3$ and $y = 4$. Let the line L be perpendicular to $3x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $(\frac{1}{2}, -5)$ from the line $L$ is equal to :

Options:

(A) $\sqrt{10}$

(B) $2\sqrt{5}$

(C) $2\sqrt{10}$

(D) $3\sqrt{10}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the first quadrant of the Cartesian plane. You see a rectangle defined by the boundaries $x = 0$, $y = 0$, $x = 3$, and $y = 4$.

We are tasked with finding a line $L$ that cuts this rectangle into two equal areas. Any line that bisects the area of a rectangle must pass through its geometric center, as this point acts as the center of symmetry.

To find this center, we calculate the midpoint of the diagonals. With vertices at $(0,0)$ and $(3,4)$, the center $C$ is the average of the coordinates:

$$C = \left( \frac{0+3}{2}, \frac{0+4}{2} \right) = \left( \frac{3}{2}, 2 \right)$$

This point $C$ serves as our anchor for the line $L$.

### The Dance of Slopes

Now, let us determine the slope of line $L$. We are given that $L$ is perpendicular to the line $3x + y + 6 = 0$.

By rearranging the reference line into the slope-intercept form $y = mx + c$, we get $y = -3x - 6$. The slope $m_1$ is $-3$.

For our line $L$ to be perpendicular, its slope $m_L$ must satisfy the condition $m_1 \cdot m_L = -1$. Substituting our known slope:

$$-3 \cdot m_L = -1 \Rightarrow m_L = \frac{1}{3}$$

We now have the point $C(\frac{3}{2}, 2)$ and the slope $m_L = \frac{1}{3}$. Using the point-slope form $y - y_1 = m(x - x_1)$, we construct the equation:

$$y - 2 = \frac{1}{3} \left( x - \frac{3}{2} \right)$$

Multiplying by $6$ to clear the fractions, we get $6(y - 2) = 2(x - \frac{3}{2})$, which simplifies to $6y - 12 = 2x - 3$. Rearranging into the general form, we arrive at the equation for line $L$:

$$2x - 6y + 9 = 0$$

### The Final Stretch: Measuring the Distance

The problem concludes by asking for the perpendicular distance from the point $P(\frac{1}{2}, -5)$ to our line $L$. We employ the perpendicular distance formula:

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

Here, $A = 2$, $B = -6$, and $C = 9$. Substituting the coordinates of $P(\frac{1}{2}, -5)$ into the numerator:

$$|2(\frac{1}{2}) - 6(-5) + 9| = |1 + 30 + 9| = 40$$

The denominator is calculated as:

$$\sqrt{2^2 + (-6)^2} = \sqrt{4 + 36} = \sqrt{40}$$

Thus, the distance $d$ is:

$$d = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$$

The final answer is **$2\sqrt{10}$**.

Select Answer:

Visualized Solution

Visualizing the Rectangle

Area Bisection Theorem

Finding the Center

Slope of the Given Line

Slope of Line $L$

Equation of Line $L$ (Setup)

Equation of Line $L$ (Compute)

The Target Point

Distance Formula Setup

Distance Formula Compute (Numerator)

Distance Formula Compute (Denominator)

Final Simplification

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Dance of Slopes

The Final Stretch

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