JEE Mathematics 2026 Question with Solution

Given: Point A lies between parallel lines $L_1$ and $L_2$. Its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Points B and C lie on $L_1$ and $L_2$, and $\triangle ABC$ is equilateral.

Find: The area of $\triangle ABC$.

Set up coordinates with $A=(0,0)$, $L_1: y=6$ and $L_2: y=-3$. Then

$$B=(x_B,6), \quad C=(x_C,-3)$$

Since the triangle is equilateral, rotating point $C$ about $A$ by $60^\circ$ sends it to $B$.

Using rotation formulas,

$$x_B=x_C\cos 60^\circ-(-3)\sin 60^\circ=x_C\left(\frac{1}{2}\right)+3\left(\frac{\sqrt{3}}{2}\right)$$

and

$$6=x_C\sin 60^\circ+(-3)\cos 60^\circ=x_C\left(\frac{\sqrt{3}}{2}\right)-3\left(\frac{1}{2}\right)$$

Now solve the second equation:

$$6=\frac{x_C\sqrt{3}-3}{2}$$ $$12=x_C\sqrt{3}-3$$ $$15=x_C\sqrt{3}$$ $$x_C=\frac{15}{\sqrt{3}}=5\sqrt{3}$$

So,

$$C=(5\sqrt{3},-3)$$

Let the side length be $s$. Then

$$s^2=AC^2=(5\sqrt{3})^2+(-3)^2=75+9=84$$

Area of an equilateral triangle is

$$\frac{\sqrt{3}}{4}s^2$$

Hence,

$$\text{Area}=\frac{\sqrt{3}}{4}\times 84=21\sqrt{3}$$

Therefore, the area of the equilateral triangle is $21\sqrt{3}$ sq. units, so the correct option is A.

Given: $A=(0,0)$, $L_1: y=6$, $L_2: y=-3$, with $B\in L_1$ and $C\in L_2$.

Find: Area of the equilateral triangle.

The hint in the solution suggests using a rotation because equilateral triangles are naturally linked with a $60^\circ$ turn. Since $AB=AC$ and $\angle BAC=60^\circ$, rotating $C$ around $A$ by $60^\circ$ places it at $B$.

Write

$$C=(x_C,-3)$$

After rotation by $60^\circ$,

$$\begin{aligned} x_B &= x_C\cos 60^\circ-(-3)\sin 60^\circ \\ 6 &= x_C\sin 60^\circ+(-3)\cos 60^\circ \end{aligned}$$

Substitute $\cos 60^\circ=\frac{1}{2}$ and $\sin 60^\circ=\frac{\sqrt{3}}{2}$:

$$6=x_C\left(\frac{\sqrt{3}}{2}\right)-\frac{3}{2}$$

Multiply by $2$:

$$12=x_C\sqrt{3}-3$$

So,

$$x_C\sqrt{3}=15$$ $$x_C=5\sqrt{3}$$

Now compute the side length using $AC$:

$$AC^2=(5\sqrt{3})^2+(-3)^2=75+9=84$$

Thus,

$$s^2=84$$

Area of an equilateral triangle:

$$\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}\cdot 84=21\sqrt{3}$$

Therefore, the correct option is A.