Q44 / 75 · JEE Main 2026 (23 January Shift 1) · Mathematics
A rectangle is formed by the lines $x=0, y=0, x=3$ and $y=4$. Let the line L be perpendicular to $3 x+y+6=0$ and divide the area of the rectangle into two equal parts. Then the distance of the point $\left(\frac{1}{2},-5\right)$ from the line $L$ is equal to :
- A$\sqrt{10}$
- B$2 \sqrt{5}$
- Correct answerC$2 \sqrt{10}$
- D$3 \sqrt{10}$
Solution
Rectangle: $x=0, y=0, x=3, y=4$, area $= 12$. Center $= (3/2, 2)$.
Slope of $3x+y+6=0$ is $-3$, so slope of $L = 1/3$.
A line through the center of a rectangle bisects its area. So $L$ passes through $(3/2, 2)$ with slope $1/3$:
$y - 2 = \frac{1}{3}(x - 3/2) $
$\Rightarrow x - 3y + \frac{9}{2} = 0$
$\Rightarrow 2x - 6y + 9 = 0$.
Distance from $(1/2, -5)$:
$d = \frac{|2(1/2) - 6(-5) + 9|}{\sqrt{4+36}}$
$= \frac{|1+30+9|}{\sqrt{40}} = \frac{40}{2\sqrt{10}} = 2\sqrt{10}$.
Slope of $3x+y+6=0$ is $-3$, so slope of $L = 1/3$.
A line through the center of a rectangle bisects its area. So $L$ passes through $(3/2, 2)$ with slope $1/3$:
$y - 2 = \frac{1}{3}(x - 3/2) $
$\Rightarrow x - 3y + \frac{9}{2} = 0$
$\Rightarrow 2x - 6y + 9 = 0$.
Distance from $(1/2, -5)$:
$d = \frac{|2(1/2) - 6(-5) + 9|}{\sqrt{4+36}}$
$= \frac{|1+30+9|}{\sqrt{40}} = \frac{40}{2\sqrt{10}} = 2\sqrt{10}$.
Hence, the correct answer is option: 'C'.