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 :
A) $\sqrt{10}$ B) $2\sqrt{5}$ C) $2\sqrt{10}$ D) $3\sqrt{10}$
A) $\sqrt{10}$ B) $2\sqrt{5}$ C) $2\sqrt{10}$ D) $3\sqrt{10}$
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Solution Steps
1
Find the Center of the Rectangle
The rectangle is bounded by $x=0$, $x=3$, $y=0$, and $y=4$. The vertices are $(0,0), (3,0), (3,4),$ and $(0,4)$.
The center of any rectangle is the midpoint of its diagonals. Center $C = (\frac{0+3}{2}, \frac{0+4}{2}) = (\frac{3}{2}, 2)$.
Property: Any line that bisects the area of a rectangle must pass through its center.
2
Determine the Slope of Line L
The given line is $3x + y + 6 = 0$. Its slope $m_1 = -3$.
Since line $L$ is perpendicular to this line, the slope of $L$ ($m_L$) satisfies $m_1 \cdot m_L = -1$.
$-3 \cdot m_L = -1 \implies m_L = \frac{1}{3}$.
3
Find the Equation of Line L
Using the point-slope form $y – y_1 = m(x – x_1)$ with point $(\frac{3}{2}, 2)$ and $m = \frac{1}{3}$:
$y – 2 = \frac{1}{3}(x – \frac{3}{2})$
$3(y – 2) = x – \frac{3}{2} \implies 3y – 6 = x – 1.5$
$x – 3y + 4.5 = 0$
Multiply by 2 to clear decimals: $2x – 6y + 9 = 0$.
4
Calculate the Distance from the Point
We need the distance of $P(\frac{1}{2}, -5)$ from $2x – 6y + 9 = 0$.
Using $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$:
$d = \frac{|2(\frac{1}{2}) – 6(-5) + 9|}{\sqrt{2^2 + (-6)^2}} = \frac{|1 + 30 + 9|}{\sqrt{4 + 36}} = \frac{40}{\sqrt{40}}$
5
Simplify the Final Result
$d = \frac{40}{\sqrt{40}} = \sqrt{40}$.
$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$.
Answer: 2√10 (Option C)
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Theory
1. Centroidal Property of Area Bisectors
A fundamental principle in Euclidean geometry is that any line passing through the geometric center (centroid) of a centrally symmetric figure, such as a circle, square, or rectangle, will divide the area of that figure into two equal parts. For a rectangle with vertices $(x_1, y_1)$ and $(x_2, y_2)$ at opposite ends of a diagonal, the center is always $( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )$. This property simplifies area-bisecting problems by converting an integration-based area problem into a simple point-on-line coordinate geometry problem.
2. Perpendicularity and Slope Relations
Two lines in a 2D plane are perpendicular if the product of their slopes is $-1$ (given the slopes are defined). If the equation of a line is provided in general form $Ax + By + C = 0$, its slope is $-A/B$. Consequently, any line perpendicular to it will have a slope of $B/A$. This relationship is a cornerstone of coordinate geometry, allowing us to find the orientation of a secondary line based solely on the coefficients of a primary reference line.
3. Point-Line Distance Derivation
The perpendicular distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ represents the shortest path between them. The formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ is derived using the vector projection of a segment connecting the point to any arbitrary point on the line onto the line’s normal vector $\vec{n} = (A, B)$. In JEE problems, this formula often acts as the final “valuation” step after determining the equation of a specific line through other geometric constraints.
4. Geometry of Rectilinear Regions
Rectangles formed by axes-parallel lines (like $x=a$ and $y=b$) are common in introductory and intermediate calculus and geometry. Because the sides are vertical and horizontal, the calculation of the area ($Area = length \times width$) and the coordinates of the center become trivial. However, the complexity increases when these shapes interact with general linear equations, requiring a bridge between simple geometric properties and algebraic manipulation of linear functions.
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FAQs
1
Why must the line pass through the center to bisect the area?
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Because a rectangle has point symmetry about its center. Any line through the center creates two congruent shapes (usually two trapezoids or two triangles), which naturally have equal areas.
2
Does this property apply to triangles?
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No. For a triangle, a line bisecting the area does not necessarily pass through the centroid. Area bisection in triangles is more complex and depends on the base and height ratios.
3
How do I find the slope of 3x + y + 6 = 0?
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Rewrite it in y = mx + c form: y = -3x – 6. The coefficient of x, which is -3, is the slope.
4
What if the line was parallel instead of perpendicular?
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If line L was parallel to 3x + y + 6 = 0, its slope would have been exactly -3.
5
How do you simplify 40/√40?
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Rationalize the denominator: (40 × √40) / (√40 × √40) = 40√40 / 40 = √40.
6
Can I use 1.5 instead of 3/2 in calculations?
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Yes, but using fractions (3/2) often makes it easier to keep the coefficients as integers, which is safer for long calculations.
7
Is the area bisector line unique?
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No, there are infinitely many lines that can bisect the area (any line through the center). However, once the slope is fixed (by the perpendicular condition), the line becomes unique.
8
What is the distance formula for the origin?
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For the origin (0,0), the distance is simply |C| / √(A² + B²).
9
What does perpendicular distance represent?
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It represents the shortest possible distance between a point and a line.
10
Why did we multiply the equation by 2?
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To convert x – 3y + 4.5 = 0 into 2x – 6y + 9 = 0. Using integer coefficients makes applying the distance formula easier and less prone to arithmetic errors.
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