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 3
Q51jee_main_2025_07_april_eveningOrthocentre of a Triangle
If the orthocentre of the triangle formed by the lines y = x + 1$y = x + 1$, y = 4x - 8$y = 4x - 8$ and y = mx + c$y = mx + c$ is at (3, -1)$(3, -1)$, then m - c$m - c$ is:
A.0$0$
B.-2$-2$
C.4$4$
D.2$2$
Solution
### Related Formula
The product of slopes of two mutually perpendicular lines is always equal to -1$-1$:
m_1 cdot m_2 = -1$$m_1 \cdot m_2 = -1$$
### Core Logic
Let the vertices of the triangle be P$P$, Q$Q$, and R$R$. The lines given are:
1) y = x + 1$y = x + 1$
2) y = 4x - 8$y = 4x - 8$
3) y = mx + c$y = mx + c$
Solving lines y = x + 1$y = x + 1$ and y = 4x - 8$y = 4x - 8$ gives the vertex P(3, 4)$P(3, 4)$.
The orthocentre is given as H(3, -1)$H(3, -1)$. Notice that the x$x$-coordinate of P$P$ and H$H$ are identical (x = 3$x = 3$). This implies that the altitude from vertex P$P$ to the base line y = mx + c$y = mx + c$ is a vertical line along x = 3$x = 3$.
Orthocentre of a Triangle diagram for Q51 - JEE Main 2025 Evening
### Step 1: Determine the Slopes
Since the altitude from P$P$ is vertical, the side opposite to it (which lies on y = mx + c$y = mx + c$) must be a horizontal line.
Therefore, the slope of the line y = mx + c$y = mx + c$ must be zero:
m = 0$m = 0$
### Step 2: Solve for c
Let's find point Q$Q$ by intersecting y = x + 1$y = x + 1$ and y = mx + c$y = mx + c$. Since m = 0$m = 0$, y = c$y = c$, we get Q(c-1, c)$Q(c-1, c)$.
Using the property that the line segment connecting Q$Q$ to the opposite side's altitude is perpendicular to line PR$PR$ (y = 4x - 8$y = 4x - 8$):
textSlope of QH cdot textSlope of PR = -1$$\text{Slope of } QH \cdot \text{Slope of } PR = -1$$frac-1 - c3 - (c - 1) cdot 4 = -1$$\frac{-1 - c}{3 - (c - 1)} \cdot 4 = -1$$frac-4(c + 1)4 - c = -1 implies 4c + 4 = 4 - c implies 5c = 0 implies c = 0$$\frac{-4(c + 1)}{4 - c} = -1 \implies 4c + 4 = 4 - c \implies 5c = 0 \implies c = 0$$
### Step 3: Evaluate m - c
Substituting the values of m$m$ and c$c$:
m - c = 0 - 0 = 0$$m - c = 0 - 0 = 0$$
### Pattern Recognition
When the x$x$-coordinate of a vertex matches the x$x$-coordinate of the orthocentre, the altitude is vertical, forcing the opposite base to be purely horizontal (m=0$m=0$). This observation cuts down calculation time completely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q69jee_main_2025_24_jan_morningConcurrency of Straight Lines
Let the lines 3x - 4y - alpha = 0$3x - 4y - \alpha = 0$, 8x - 11y - 33 = 0$8x - 11y - 33 = 0$, and 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$ be concurrent. If the image of the point (1, 2)$(1, 2)$ in the line 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$ is left(frac5713,frac-4013
ight)$\left(\frac{57}{13},\frac{-40}{13}
ight)$ , then |alpha lambda|$|\alpha \lambda|$ is equal to :
A.84$84$
B.91$91$
C.113$113$
D.101$101$
Solution
### Related Formula
The midpoint between a point and its reflection image must lie exactly on the line mirror equation.
### Core Logic
Find the midpoint M$M$ between point P(1, 2)$P(1, 2)$ and its given reflection image Qleft(frac5713, frac-4013right)$Q\left(\frac{57}{13}, \frac{-40}{13}\right)$:
M = left( frac1 + frac57132, \, frac2 - frac40132 right) = left( frac7026, \, frac-1426 right) = left( frac3513, \, frac-713 right)$$M = \left( \frac{1 + \frac{57}{13}}{2}, \, \frac{2 - \frac{40}{13}}{2} \right) = \left( \frac{70}{26}, \, \frac{-14}{26} \right) = \left( \frac{35}{13}, \, \frac{-7}{13} \right)$$
Since M$M$ lies on the reflecting line 2x - 3y + lambda = 0$2x - 3y + \lambda = 0$:
2left(frac3513right) - 3left(frac-713right) + lambda = 0$$2\left(\frac{35}{13}\right) - 3\left(\frac{-7}{13}\right) + \lambda = 0$$frac7013 + frac2113 + lambda = 0 implies frac9113 + lambda = 0 implies 7 + lambda = 0 implies lambda = -7$$\frac{70}{13} + \frac{21}{13} + \lambda = 0 \implies \frac{91}{13} + \lambda = 0 \implies 7 + \lambda = 0 \implies \lambda = -7$$
### Step 1: Apply Concurrency Determinant
For three straight lines to intersect at a single concurrent point, the determinant of their linear coefficients must equal zero:
left| beginmatrix 3 & -4 & -alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 endmatrix right| = 0$$\left| \begin{matrix} 3 & -4 & -\alpha \\ 8 & -11 & -33 \\ 2 & -3 & -7 \end{matrix} \right| = 0$$
Expand the determinant along the first row:
3left[ (-11)(-7) - (-33)(-3) right] - (-4)left[ (8)(-7) - (-33)(2) right] - alpha left[ (8)(-3) - (-11)(2) right] = 0$$3\left[ (-11)(-7) - (-33)(-3) \right] - (-4)\left[ (8)(-7) - (-33)(2) \right] - \alpha \left[ (8)(-3) - (-11)(2) \right] = 0$$3[77 - 99] + 4[-56 + 66] - alpha[-24 + 22] = 0$$3[77 - 99] + 4[-56 + 66] - \alpha[-24 + 22] = 0$$3[-22] + 4[10] - alpha[-2] = 0$$3[-22] + 4[10] - \alpha[-2] = 0$$-66 + 40 + 2alpha = 0 implies 2alpha = 26 implies alpha = 13$$-66 + 40 + 2\alpha = 0 \implies 2\alpha = 26 \implies \alpha = 13$$
### Step 2: Compute Final Product Target
Multiply the absolute values of the determined parameters together:
|alpha lambda| = |13 cdot (-7)| = |-91| = 91$$|\alpha \lambda| = |13 \cdot (-7)| = |-91| = 91$$
### Pattern Recognition
Using the midpoint property to evaluate unknown line parameters from reflection images is often much faster than using full distance formulas.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q67jee_main_2025_28_jan_eveningAngle Between Lines
Two equal sides of an isosceles triangle are along -x+2y=4$-x+2y=4$ and x+y=4.$x+y=4.$ If m is the slope of its third side, then the sum, of all possible distinct values of m, is:
A.-6$-6$
B.12$12$
C.6$6$
D.-2sqrt10$-2\sqrt{10}$
Solution
### Related Formula
Angle theta$\theta$ between two lines with slopes m_1$m_1$ and m_2$m_2$:
tantheta = left| fracm_1 - m_21 + m_1 m_2 right|$$\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$
### Core Logic
Given lines for the equal sides:
1) -x + 2y = 4 implies y = frac12x + 2 implies m_1 = frac12$-x + 2y = 4 \implies y = \frac{1}{2}x + 2 \implies m_1 = \frac{1}{2}$
2) x + y = 4 implies y = -x + 4 implies m_2 = -1$x + y = 4 \implies y = -x + 4 \implies m_2 = -1$
In an isosceles triangle, the third side makes equal angles theta$\theta$ with both equal sides. Let the slope of the third side be m$m$:
left| fracm - 1/21 + m/2 right| = left| fracm - (-1)1 + m(-1) right|$$\left| \frac{m - 1/2}{1 + m/2} \right| = \left| \frac{m - (-1)}{1 + m(-1)} \right|$$left| frac2m - 12 + m right| = left| fracm + 11 - m right|$$\left| \frac{2m - 1}{2 + m} \right| = \left| \frac{m + 1}{1 - m} \right|$$
### Step 1: Solve the Slope Equation
Case 1 (Same sign):
frac2m - 12 + m = fracm + 11 - m$$\frac{2m - 1}{2 + m} = \frac{m + 1}{1 - m}$$(2m - 1)(1 - m) = (m + 1)(2 + m)$$(2m - 1)(1 - m) = (m + 1)(2 + m)$$2m - 2m^2 - 1 + m = m^2 + 3m + 2$$2m - 2m^2 - 1 + m = m^2 + 3m + 2$$-2m^2 + 3m - 1 = m^2 + 3m + 2$$-2m^2 + 3m - 1 = m^2 + 3m + 2$$3m^2 + 3 = 0 implies m^2 = -1 quad (textNo real roots)$$3m^2 + 3 = 0 \implies m^2 = -1 \quad (\text{No real roots})$$
Case 2 (Opposite sign):
frac2m - 12 + m = -fracm + 11 - m = fracm + 1m - 1$$\frac{2m - 1}{2 + m} = -\frac{m + 1}{1 - m} = \frac{m + 1}{m - 1}$$(2m - 1)(m - 1) = (2 + m)(m + 1)$$(2m - 1)(m - 1) = (2 + m)(m + 1)$$2m^2 - 3m + 1 = m^2 + 3m + 2$$2m^2 - 3m + 1 = m^2 + 3m + 2$$m^2 - 6m - 1 = 0$$m^2 - 6m - 1 = 0$$
### Step 2: Sum of Roots
The quadratic equation for m$m$ is m^2 - 6m - 1 = 0$m^2 - 6m - 1 = 0$.
The sum of possible distinct values of m$m$ is given by the sum of roots of this quadratic:
textSum of roots = -frac-61 = 6$$\text{Sum of roots} = -\frac{-6}{1} = 6$$
### Pattern Recognition
Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation m^2 - 6m - 1 = 0$m^2 - 6m - 1 = 0$ gives the final answer instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q54jee_main_2025_29_jan_morningCentroid and Image of a Point
Let ABC be a triangle formed by the lines 7mathrmx - 6mathrmy + 3 = 0, mathrmx + 2mathrmy - 31 = 0$7\mathrm{x} - 6\mathrm{y} + 3 = 0, \mathrm{x} + 2\mathrm{y} - 31 = 0$ and 9mathrmx - 2mathrmy - 19 = 0$9\mathrm{x} - 2\mathrm{y} - 19 = 0$ . Let the point (mathrmh,mathrmk)$(\mathrm{h},\mathrm{k})$ be the image of the centroid of Delta ABC$\Delta ABC$ in the line 3mathrmx + 6mathrmy - 53 = 0$3\mathrm{x} + 6\mathrm{y} - 53 = 0$ . Then \mathrm{h}^2 + \mathrm{k}^2 + \mathrm{hk} is equal to
A. 37
B. 47
C. 40
D. 36
Solution
### Related Formula
textCentroid G = left(fracx_1+x_2+x_33, fracy_1+y_2+y_33right)$$\text{Centroid } G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$textImage of (x_1, y_1) text in line ax+by+c=0: fracx-x_1a = fracy-y_1b = -2fracax_1+by_1+ca^2+b^2$$\text{Image of } (x_1, y_1) \text{ in line } ax+by+c=0: \frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$$
### Core Logic
First, find the vertices A, B, C$A, B, C$ by solving the lines pairwise.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and x + 2y - 31 = 0$x + 2y - 31 = 0$ gives A(9,11)$A(9,11)$.
Solving 7x - 6y + 3 = 0$7x - 6y + 3 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives B(3,4)$B(3,4)$.
Solving x + 2y - 31 = 0$x + 2y - 31 = 0$ and 9x - 2y - 19 = 0$9x - 2y - 19 = 0$ gives C(5,13)$C(5,13)$.
Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 1: Determine the Centroid
G = left(frac9 + 3 + 53, frac11 + 4 + 133right) = left(frac173, frac283right)$$G = \left(\frac{9 + 3 + 5}{3}, \frac{11 + 4 + 13}{3}\right) = \left(\frac{17}{3}, \frac{28}{3}\right)$$
### Step 2: Find the Image (h, k)
Using the line 3x + 6y - 53 = 0$3x + 6y - 53 = 0$:
frach - frac1733 = frack - frac2836 = -2 frac3left(frac173right) + 6left(frac283right) - 533^2 + 6^2$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{3\left(\frac{17}{3}\right) + 6\left(\frac{28}{3}\right) - 53}{3^2 + 6^2}$$frach - frac1733 = frack - frac2836 = -2 frac17 + 56 - 5345 = -2 frac2045 = -frac89$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = -2 \frac{17 + 56 - 53}{45} = -2 \frac{20}{45} = -\frac{8}{9}$$
Solving for h$h$ and k$k$ yields:
h = 3, quad k = 4$$h = 3, \quad k = 4$$Centroid diagram for Q54 - JEE Main 2025 Morning
### Step 3: Compute final algebraic target
h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$h^2 + k^2 + hk = 3^2 + 4^2 + (3)(4) = 9 + 16 + 12 = 37$$
### Pattern Recognition
Instead of solving fractions endlessly, substitute potential integer coordinates early into the slope relationship (k - y_G)/(h - x_G) = -1/m$(k - y_G)/(h - x_G) = -1/m$ to accelerate competitive solving time.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q11jee_main_2024_29_january_eveningDistance of a Point From a Line
The distance of the point (2, 3) from the line 2x - 3y + 28 = 0$2x - 3y + 28 = 0$, measured parallel to the line sqrt3 x - y + 1 = 0$sqrt{3} x - y + 1 = 0$, is equal to
A.4sqrt2$4\sqrt{2}$
B.6sqrt3$6\sqrt{3}$
C.3 + 4sqrt2$3 + 4\sqrt{2}$
D.4 + 6sqrt3$4 + 6\sqrt{3}$
Solution
### Related Formula
x = x_1 + r cos theta, quad y = y_1 + r sin theta$$x = x_1 + r \cos \theta, \quad y = y_1 + r \sin \theta$$
### Core Logic
The line is measured parallel to sqrt3x - y + 1 = 0$\sqrt{3}x - y + 1 = 0$, which has a slope tan theta = sqrt3 implies theta = 60^circ$\tan \theta = \sqrt{3} \implies \theta = 60^\circ$.
Thus, cos theta = frac12$\cos \theta = \frac{1}{2}$ and \sin \theta = \frac{\sqrt{3}}{2}.
Writing any point $.
Writing any point $P along this direction passing through $ along this direction passing through $(2,3) in parametric coordinates:
$ in parametric coordinates:
$P = left(2 + r cos 60^circ, 3 + r sin 60^circright) = left(2 + fracr2, 3 + fracsqrt3r2right)$P = \left(2 + r \cos 60^\circ, 3 + r \sin 60^\circ\right) = \left(2 + \frac{r}{2}, 3 + \frac{\sqrt{3}r}{2}\right)$
### Step 1: Finding Intersection Point
Since $
### Step 1: Finding Intersection Point
Since $P must lie on the given line $ must lie on the given line $2x - 3y + 28 = 0:
$:
$2left(2 + fracr2right) - 3left(3 + fracsqrt3r2right) + 28 = 0$2\left(2 + \frac{r}{2}\right) - 3\left(3 + \frac{\sqrt{3}r}{2}\right) + 28 = 0$$
$4 + r - 9 - frac3sqrt3r2 + 28 = 0$4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$$
$23 + rleft(1 - frac3sqrt32right) = 0$23 + r\left(1 - \frac{3\sqrt{3}}{2}\right) = 0$$
$rleft(frac3sqrt3 - 22right) = 23 implies r = frac463sqrt3 - 2$r\left(\frac{3\sqrt{3} - 2}{2}\right) = 23 \implies r = \frac{46}{3\sqrt{3} - 2}$
Rationalizing the denominator:
$
Rationalizing the denominator:
$r = frac46(3sqrt3 + 2)(3sqrt3)^2 - 2^2 = frac46(3sqrt3 + 2)27 - 4 = frac46(3sqrt3 + 2)23 = 2(3sqrt3 + 2) = 4 + 6sqrt3$r = \frac{46(3\sqrt{3} + 2)}{(3\sqrt{3})^2 - 2^2} = \frac{46(3\sqrt{3} + 2)}{27 - 4} = \frac{46(3\sqrt{3} + 2)}{23} = 2(3\sqrt{3} + 2) = 4 + 6\sqrt{3}$$
### Pattern Recognition
Distance measured parallel to a given direction is always resolved most efficiently using parametric equations of lines rather than perpendicular metrics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: 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.
