Let a point A lie between the parallel lines L_1$L_{1}$ and L_2$L_{2}$ such that its distances from L_1$L_{1}$ and L_2$L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L_1$L_{1}$ and L_2$L_{2}$ respectively, is:
A.15sqrt6$15\sqrt{6}$
B.27
C.21sqrt3$21\sqrt{3}$
D.12sqrt2$12\sqrt{2}$
Solution & Explanation
### Related Formula
textArea of Equilateral Triangle = fracsqrt34 a^2$$\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2$$
Where a$a$ is the side length.
### Core Logic
Let the side of the equilateral triangle be a$a$.
Let theta$\theta$ be the angle between the side AC$AC$ and the parallel line L_2$L_2$.
Then, the angle between the side AB$AB$ and the parallel line L_1$L_1$ can be expressed via alternate geometry. Given distances from A$A$ to the lines form right-angled triangles.
### Step 1: Set up geometric projections
Equilateral triangle between parallel lines diagram for Q4 - JEE Main 2026 Morning
From vertex C$C$ to line passing through A$A$ parallel to L_1, L_2$L_1, L_2$, the perpendicular distance is 3$3$.
In the right triangle formed, we have:
sin theta = frac3a$$\sin \theta = \frac{3}{a}$$
Similarly, point B$B$ lies on L_1$L_1$. The perpendicular distance from A$A$ to L_1$L_1$ is 6$6$.
However, combining the overall heights between the parallel lines, the total distance between L_1$L_1$ and L_2$L_2$ is 6 + 3 = 9$6 + 3 = 9$.
The projection of side BC$BC$ (which connects L_1$L_1$ and L_2$L_2$) vertically is 9$9$.
sin(60^circ + theta) = frac9a$$\sin(60^{\circ} + \theta) = \frac{9}{a}$$
### Step 2: Solve the trigonometric system
Expand sin(60^circ + theta)$\sin(60^{\circ} + \theta)$:
fracsqrt32 cos theta + frac12 sin theta = frac9a$$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$$
Substitute sin theta = frac3a$\sin \theta = \frac{3}{a}$ and cos theta = sqrt1 - sin^2 theta = sqrt1 - frac9a^2$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{a^2}}$:
fracsqrt32 sqrt1 - frac9a^2 + frac12 left(frac3aright) = frac9a$$\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \left(\frac{3}{a}\right) = \frac{9}{a}$$sqrt3 sqrt1 - frac9a^2 + frac3a = frac18a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$$sqrt3 sqrt1 - frac9a^2 = frac15a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} = \frac{15}{a}$$
Squaring both sides:
3 left(1 - frac9a^2right) = frac225a^2$$3 \left(1 - \frac{9}{a^2}\right) = \frac{225}{a^2}$$3 - frac27a^2 = frac225a^2$$3 - \frac{27}{a^2} = \frac{225}{a^2}$$3 = frac252a^2 Rightarrow a^2 = 84$$3 = \frac{252}{a^2} \Rightarrow a^2 = 84$$
### Step 3: Calculate Area
textArea of Delta ABC = fracsqrt34 a^2$$\text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} a^2$$= fracsqrt34 times 84 = 21sqrt3$$= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$$
### Pattern Recognition
When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle theta$\theta$ for one edge and use rotational shifts (e.g., 60^circ + theta$60^{\circ} + \theta$) to project heights. Expanding the sine addition formula instantly yields the side length.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Class 11 Maths: Trigonometric Functions
Keywords:#area of the equilateral triangle#JEE Main 2026 Morning Q4#Straight Lines JEE Main 2026#Equilateral Triangle Between Parallel Lines JEE Main 2026
More Straight Lines Previous-Year Questions — Page 5
Q2jee_main_2024_31_jan_eveningCentroid and Orthocentre
Let A (a, b)$A (a, b)$, B(3, 4)$B(3, 4)$ and (-6, -8)$(-6, -8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5)$P(2a + 3, 7b + 5)$ from the line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$ measured parallel to the line x - 2y - 1 = 0$x - 2y - 1 = 0$ is
A.frac15 sqrt57$\frac{15 \sqrt{5}}{7}$
B.frac17sqrt56$\frac{17\sqrt{5}}{6}$
C.frac17 sqrt57$\frac{17 \sqrt{5}}{7}$
D.fracsqrt517$\frac{\sqrt{5}}{17}$
Solution
### Related Formula
textCentroid divides the line joining Orthocentre and Circumcentre in 2:1$$\text{Centroid divides the line joining Orthocentre and Circumcentre in } 2:1$$
Distance in parametric form: x = x_1 + rcostheta, y = y_1 + rsintheta$x = x_1 + r\cos\theta, y = y_1 + r\sin\theta$
### Core Logic
Centroid and Orthocentre diagram for Q2 - JEE Main 2024 Evening
Let Orthocentre C(-6, -8)$C(-6, -8)$ and Circumcentre B(3, 4)$B(3, 4)$. Centroid A(a, b)$A(a, b)$ divides CB$CB$ in 2:1$2:1$.
a = frac2(3) + 1(-6)2+1 = 0$$a = \frac{2(3) + 1(-6)}{2+1} = 0$$b = frac2(4) + 1(-8)2+1 = 0$$b = \frac{2(4) + 1(-8)}{2+1} = 0$$
So, P(2a+3, 7b+5) = (3, 5)$P(2a+3, 7b+5) = (3, 5)$.
The line along which distance is measured is parallel to x - 2y - 1 = 0$x - 2y - 1 = 0$, giving slope m = tantheta = frac12$m = \tan\theta = \frac{1}{2}$.
Using parametric coordinates from P(3,5)$P(3,5)$:
x = 3 + rcostheta, quad y = 5 + rsintheta$$x = 3 + r\cos\theta, \quad y = 5 + r\sin\theta$$
Substitute into the target line 2x + 3y - 4 = 0$2x + 3y - 4 = 0$:
2(3 + rcostheta) + 3(5 + rsintheta) - 4 = 0$$2(3 + r\cos\theta) + 3(5 + r\sin\theta) - 4 = 0$$r(2costheta + 3sintheta) = -17$$r(2\cos\theta + 3\sin\theta) = -17$$
From tantheta = 1/2$\tan\theta = 1/2$, we get sintheta = frac1sqrt5$\sin\theta = \frac{1}{\sqrt{5}}$ and costheta = frac2sqrt5$\cos\theta = \frac{2}{\sqrt{5}}$.
rleft(2left(frac2sqrt5right) + 3left(frac1sqrt5right)right) = -17$$r\left(2\left(\frac{2}{\sqrt{5}}\right) + 3\left(\frac{1}{\sqrt{5}}\right)\right) = -17$$rleft(frac7sqrt5right) = -17 implies |r| = frac17sqrt57$$r\left(\frac{7}{\sqrt{5}}\right) = -17 \implies |r| = \frac{17\sqrt{5}}{7}$$
### Pattern Recognition
Standard Euler line property: O, G, C$O, G, C$ are collinear and G$G$ divides OC$OC$ in 2:1$2:1$. Use parametric equation to find intersection distance directly without finding the intersection point.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Let A(-2, -1)$A(-2, -1)$, B(1, 0)$B(1, 0)$, C(alpha, beta)$C(\alpha, \beta)$ and D(gamma, delta)$D(\gamma, \delta)$ be the vertices of a parallelogram ABCD. If the point C lies on 2x - y = 5$2x - y = 5$ and the point D lies on 3x - 2y = 6$3x - 2y = 6$, then the value of |alpha + beta + gamma + delta|$|\alpha + \beta + \gamma + \delta|$ is equal to
Q10jee_main_2024_31_jan_morningProperties of Parallelogram
Let alpha, beta, gamma, delta in Z$\alpha, \beta, \gamma, \delta \in Z$ and let A (alpha, beta)$A (\alpha, \beta)$, B (1, 0)$B (1, 0)$, C (gamma, delta)$C (\gamma, \delta)$ and D (1, 2)$D (1, 2)$ be the vertices of a parallelogram ABCD$ABCD$. If AB = sqrt10$AB = \sqrt{10}$ and the points A$A$ and C$C$ lie on the line 3y = 2x + 1$3y = 2x + 1$, then 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$ is equal to
A.10$10$
B.5$5$
C.12$12$
D.8$8$
Solution
### Core Logic
Let E$E$ be the midpoint of the diagonals AC$AC$ and BD$BD$.
Since ABCD$ABCD$ is a parallelogram, the diagonals bisect each other.
Properties of Parallelogram diagram for Q10 - JEE Main 2024 Morning
Midpoint E$E$ from BD$BD$: left( frac1+12, frac0+22 right) = (1, 1)$\left( \frac{1+1}{2}, \frac{0+2}{2} \right) = (1, 1)$.
### Step 1: Apply Midpoint on AC
Midpoint E$E$ from AC$AC$: left( fracalpha+gamma2, fracbeta+delta2 right)$\left( \frac{\alpha+\gamma}{2}, \frac{\beta+\delta}{2} \right)$.
Equating both:
fracalpha+gamma2 = 1 implies alpha + gamma = 2$$\frac{\alpha+\gamma}{2} = 1 \implies \alpha + \gamma = 2$$fracbeta+delta2 = 1 implies beta + delta = 2$$\frac{\beta+\delta}{2} = 1 \implies \beta + \delta = 2$$
### Step 2: Final Value
The expression requires 2(alpha + beta + gamma + delta)$2(\alpha + \beta + \gamma + \delta)$.
2(2 + 2) = 2(4) = 8$$2(2 + 2) = 2(4) = 8$$
*(Note: Additional conditions like AB = sqrt10$AB = \sqrt{10}$ and the line equation are extraneous data not needed to find the sum).*
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
More Straight Lines Questions — jee_main_2026_21_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
