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JEE Mathematics 2026 Question with Solution

A rectangle is formed by the lines x=0x = 0, y=0y = 0, x=3x = 3 and y=4y = 4. Let the line LL be perpendicular to 3x+y+6=03x + y + 6 = 0 and divide the area of the rectangle into two equal parts. Then the distance of the point (12,−5)\left(\frac{1}{2}, -5\right) from the line LL is equal to :

  • A

    252\sqrt{5}

  • B

    2102\sqrt{10}

  • C

    3103\sqrt{10}

  • D

    10\sqrt{10}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A rectangle is bounded by x=0x = 0, y=0y = 0, x=3x = 3 and y=4y = 4. The line LL is perpendicular to 3x+y+6=03x + y + 6 = 0 and divides the rectangle into two equal areas.

Find: The distance of the point (12,−5)\left(\frac{1}{2}, -5\right) from the line LL.

A line that divides the area of a rectangle into two equal parts must pass through the center of the rectangle.

The center of the rectangle is

(0+32,0+42)=(1.5,2)\left(\frac{0+3}{2}, \frac{0+4}{2}\right) = \left(1.5, 2\right)

The slope of the given line 3x+y+6=03x + y + 6 = 0 is −3-3. Therefore, the slope of the perpendicular line LL is

m=13m = \frac{1}{3}

Using the point-slope form through (1.5,2)\left(1.5, 2\right),

y−2=13(x−1.5)y - 2 = \frac{1}{3}(x - 1.5)

Rearranging,

3y−6=x−1.53y - 6 = x - 1.5 x−3y+4.5=0x - 3y + 4.5 = 0

Multiplying by 22,

2x−6y+9=02x - 6y + 9 = 0

Now the distance of (12,−5)\left(\frac{1}{2}, -5\right) from this line is

d=∣2(12)−6(−5)+922+(−6)2∣d = \left| \frac{2\left(\frac{1}{2}\right) - 6(-5) + 9}{\sqrt{2^2 + (-6)^2}} \right| d=∣1+30+940∣=4040=40=210d = \left| \frac{1 + 30 + 9}{\sqrt{40}} \right| = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}

Therefore, the distance is 2102\sqrt{10}. The correct option is B.

Common mistakes

  • Assuming any perpendicular line to 3x+y+6=03x + y + 6 = 0 will bisect the rectangle. This is wrong because the line must also pass through the center of the rectangle to divide its area into two equal parts. First find the center, then use the perpendicular slope.

  • Using the same slope −3-3 for the line LL. This is incorrect because perpendicular lines have slopes whose product is −1-1. Since the given slope is −3-3, the slope of LL must be 13\frac{1}{3}.

  • Making an error in the point-to-line distance formula by forgetting the modulus or using the wrong denominator. For a line ax+by+c=0ax + by + c = 0, the distance from (x1,y1)\left(x_1, y_1\right) is ∣ax1+by1+ca2+b2∣\left|\frac{ax_1 + by_1 + c}{\sqrt{a^2+b^2}}\right|.

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