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MCQMediumJEE 2024Conic Sections (Parabola, Ellipse, Hyperbola)

JEE Mathematics 2024 Question with Solution

The vertices of a triangle are A(−1,3-1, 3), B(−2,2-2, 2), and C(3,−13, -1). A new triangle is formed by shifting the sides of the triangle one unit inwards. Then the equation of the side of the new triangle nearest to the origin is:

  • A

    x−y−(2+2)=0x - y - (2 + \sqrt{2}) = 0

  • B

    −x+y−(2−2)=0-x + y - (2 - \sqrt{2}) = 0

  • C

    x+y−(2−2)=0x + y - (2 - \sqrt{2}) = 0

  • D

    x+y+(2−2)=0x + y + (2 - \sqrt{2}) = 0

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The triangle has vertices A(−1,3-1, 3), B(−2,2-2, 2), and C(3,−13, -1). The sides are shifted inward by 11 unit.

Find: The equation of the side of the new triangle nearest to the origin.

From the given solution, the relevant side is AC. Its equation is

x+y=2x + y = 2

A line parallel to x+y=2x + y = 2 has the form

x+y=kx + y = k

The distance between the two parallel lines x+y−2=0x + y - 2 = 0 and x+y−k=0x + y - k = 0 is

∣k−2∣12+12=∣k−2∣2\frac{|k - 2|}{\sqrt{1^2 + 1^2}} = \frac{|k - 2|}{\sqrt{2}}

Since the side is shifted inward by 11 unit,

∣k−2∣2=1\frac{|k - 2|}{\sqrt{2}} = 1

So,

∣k−2∣=2|k - 2| = \sqrt{2}

For the inward shift toward the origin, we take

k=2−2k = 2 - \sqrt{2}

Hence the required side is

x+y=2−2x + y = 2 - \sqrt{2}

or equivalently,

x+y−(2−2)=0x + y - (2 - \sqrt{2}) = 0

Therefore, the correct option is C.

Common mistakes

  • Using the wrong side of the triangle. The solution specifically identifies side AC with equation x+y=2x + y = 2. Starting from another side leads to a different parallel line.

  • Forgetting the distance formula for parallel lines. The shift is 11 unit, but the constant term does not change by 11 directly because the normal vector length is 2\sqrt{2}. Use ∣c1−c2∣a2+b2\frac{|c_1-c_2|}{\sqrt{a^2+b^2}} instead.

  • Choosing 2+22 + \sqrt{2} instead of 2−22 - \sqrt{2}. Both are parallel shifts, but only the inward shift gives the side nearer to the origin.

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