QJEEM25Ma1359: JEE Main 2025 Mathematics - Class 11 JEEM25Ma1359 Solution: Let the distance between two parallel lines be 5 units and a…

We set up a coordinate system so that the two parallel lines are given by

$y = 0 \quad \text{and} \quad y = 5,$

since their distance is 5 units. Choose point

$P = (0,1)$

so that the distance from $P$ to the line $y=0$ is 1 unit (and its distance to the line $y=5$ is 4 units).

Let point

$Q = (a,0)$

be on the line $y = 0$ , and let point

$R = (b,5)$

be on the line $y = 5$ . Since triangle $PQR$ is equilateral with side length $s$ , we require:

$PQ = PR = QR = s.$

A convenient method is to “rotate” $Q$ about $P$ by an angle of $60^\circ$ to obtain $R$ . In complex-number (or vector) terms, if we translate so that $P$ is at the origin, then the rotation is given by

$e^{i60^\circ} = \cos 60^\circ + i \sin 60^\circ = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$

Thus, writing $Q$ in vector form relative to $P$ , we have

$Q - P = (a, -1).$

Rotating this by $60^\circ$ gives

$R - P = \left(a\cos60^\circ - (-1)\sin60^\circ,\; a\sin60^\circ + (-1)\cos60^\circ\right).$

Substituting the values $\cos60^\circ = \frac{1}{2}$ and $\sin60^\circ = \frac{\sqrt{3}}{2}$ , we obtain

$\begin{aligned} R - P &= \left(\frac{a}{2} + \frac{\sqrt{3}}{2},\; \frac{a\sqrt{3}}{2} - \frac{1}{2}\right), \\ \text{so} \quad R &= \left( \frac{a+\sqrt{3}}{2},\; 1 + \frac{a\sqrt{3}}{2} - \frac{1}{2} \right) = \left( \frac{a+\sqrt{3}}{2},\; \frac{a\sqrt{3}+1}{2} \right). \end{aligned}$

Since $R$ lies on $y = 5$ , its $y$ -coordinate must equal 5:

$\frac{a\sqrt{3}+1}{2} = 5.$

Solve for $a$ :

$\begin{aligned} a\sqrt{3} + 1 &= 10, \\ a\sqrt{3} &= 9, \\ a &= \frac{9}{\sqrt{3}} = 3\sqrt{3}. \end{aligned}$

Now, the side length $s$ (which is the distance $PQ$ ) is given by

$\begin{aligned} s^2 &= PQ^2 = \left(3\sqrt{3} - 0\right)^2 + \left(0 - 1\right)^2 \\ &= (3\sqrt{3})^2 + 1^2 \\ &= 27 + 1 \\ &= 28. \end{aligned}$

Thus, the square of side $QR$ is

$(QR)^2 = s^2 = 28.$