JEE Main 2025 Apr 8 Shift 2, Mathematics Q7: Points and Straight Lines

JEE Main2025Apr 8, Shift 2Mathematics

Let $a$ be the length of a side of a square $O A B C$ with $O$ being the origin. Its side $O A$ makes an acute angle $α$ with the positive x-axis and the equations of its diagonals are $(3 + 1) x + (3 - 1) y = 0$ and $(3 - 1) x - (3 + 1) y + 83 = 0$ . Then $a^{2}$ is equal to

A
$48$
B
$32$
C
$16$
D
$24$

Solution

The two diagonals of a square meet at right angles at the centre, and each makes a $4 5^{\circ}$ angle with the sides.

The first diagonal passes through the origin with slope $- \frac{3 + 1}{3 - 1}$ , which after rationalising equals $\frac{3 + 1}{1 - 3} = tan 10 5^{\circ}$ ; this is the diagonal $O B$ .

Since $O A$ makes angle $α$ with the x-axis and the diagonal $O B$ makes angle $α + 4 5^{\circ} = 10 5^{\circ}$ , we get $α = 6 0^{\circ}$ .

Vertex $A$ then has coordinates $(a cos 6 0^{\circ}, a sin 6 0^{\circ}) = (\frac{a}{2}, \frac{a 3}{2})$ .

Point $A$ lies on the second diagonal, so substituting into $(3 - 1) x - (3 + 1) y + 83 = 0$ : $(\frac{3 - 1}{2}) a - (\frac{3 + 1}{2}) a 3 + 83 = 0$ .

Simplifying the coefficient of $a$ : $\frac{3 - 1 - 3 - 3}{2} = - 2$ , so $- 2 a = - 83$ , giving $a = 43$ .

Therefore $a^{2} = 48$ , matching option (1).

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JEE Main 2025 Apr 8 Shift 2, Mathematics Q7: Points and Straight Lines

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