JEE Advanced 2025 — Paper 1 — MATH-III solution

JEE Advanced2025Paper 1MATH-III

For any two points $M$ and $N$ in the $X Y$ -plane, let $M N$ denote the vector from $M$ to $N$ , and $0$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $X Y$ -plane. Let $S$ be a point inside the triangle $△ P QR$ such that

$S P + 5 S Q + 6 S R = 0 .$

Let $E$ and $F$ be the mid-points of the sides $P R$ and $QR$ , respectively. Then the value of

$\frac{length of the line segment E F}{length of the line segment E S}$

is _______ .

Your answer

Solution

Let position vectors of $P, Q, R, S$ be $p, q, r, s$ . The given condition rewrites as

$(p - s) + 5 (q - s) + 6 (r - s) = 0,$

which gives $s = \frac{p + 5 q + 6 r}{12} .$

Compute the two relevant displacements. Since $E$ is the midpoint of $P R$ and $F$ is the midpoint of $QR$ , by the midpoint theorem $E F = \frac{q - p}{2} .$

Also $E S = s - \frac{p + r}{2} = \frac{p + 5 q + 6 r}{12} - \frac{6 p + 6 r}{12} = \frac{- 5 p + 5 q}{12} = \frac{5 ( q - p )}{12} .$

Both displacements are parallel to $q - p$ , so

$\frac{∣ E F ∣}{∣ E S ∣} = \frac{1/2}{5/12} = \frac{12}{10} = 1.20.$

$∴$ The answer is $1.20$ .

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