Let a be the length of a side of a square OABC with O being the origin. Its side OA 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 a2 is equal to
(A)48✓ Correct
(B)32
(C)16
(D)24
Answer: (A)
Solution
The diagonal through the origin, (3+1)x+(3−1)y=0, has slope −3−13+1=−(2+3)=tan105∘. A diagonal makes 45∘ with a side of the square, so α+45∘=105∘ and α=60∘; hence A=(acos60∘,asin60∘)=(2a,23a). Since A lies on the other diagonal (3−1)x−(3+1)y+83=0, substitution gives 2a[(3−1)−3(3+1)]+83=0, that is 2a(−4)=−83, so a=43 and a2=48.
JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option