JEE Main 2026 Jan 23 Shift 1, Mathematics Q4: Points and Straight Lines

JEE Main2026Jan 23, Shift 1Mathematics

Q.

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 $(\frac{1}{2}, - 5)$ from the line L is equal to :

A
$25$
B
$310$
C
$10$
D
$210$

Solution

Slope of L. Perpendicular to $3 x + y + 6 = 0$ (slope $- 3$ ), so L has slope $\frac{1}{3}$ .

Bisecting condition. A line of slope $\frac{1}{3}$ that bisects the rectangle's area must pass through its centre $(\frac{3}{2}, 2)$ .

$L : y - 2 = \frac{1}{3} (x - \frac{3}{2}) \Rightarrow 2 x - 6 y + 9 = 0$

Distance from $(\frac{1}{2}, - 5)$ :

$d = \frac{∣2 ( \frac{1}{2} ) - 6 ( - 5 ) + 9∣}{4 + 36} = \frac{∣1 + 30 + 9∣}{40} = \frac{40}{2 10} = 210$ .

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